Two Way Slab at EL (+)3.5M

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SUNIL HITECH ENGINEERS LIMITED

Clear span of Short span = 2900 mm

Clear span of Long span = 5700 mm

Width of the Short span support = 600 mm

Width of the Long span support = 800 mm

Overall depth of the slab assumed,D = 150 mm

Clear cover to main reinforcement = 25 mm

25

1

Load Calculations:

Dead load of the slab = 3.75

Live load on the slab = 15.00

Floor Finish on the slab = 1.00

Total Load on the Slab, w = 19.75

120 mm

111 mm

2900 mm

5700 mm

1.97

Therefore It is a Two-way slab

Bending Moment Coefficients:Short-span:

0.047

0.063

Long-span:

0.024

0.032

Bending Moment Calculations:

Short span:

0.047 11.74 kN-m0.063 15.65 kN-m

Long-span:

0.024 5.98 kN-m

0.032 7.97 kN-m

Check for depth:

Effective depth required,d = =sqrt(Mu/(0.133.fck.b))= 68.60 mm

Overall depth of the Slab, D = 98.60 mm

Assumed depth is adequate.

TCE-5910A-151-DC-6857

SHEL-PRL-720-CVC-R-857

RAW WATER PUMP HOUSE.DESIGN OF TWO WAY SLAB

AT EL(+)3.5 M

4.2 Design of Two-Way Slab:- From Grid 10 TO 11

Characteristic Comp. Strength of Concrete,f

ck = N/mm2

Support Condition :

kN/m2

kN/m2

kN/m2

kN/m2

Effective depth in Short direction, dx

=

Effective depth in Long direction, dy

=

Effective Length of Short Span, lx

=

Effective Length of Long Span, ly

=

Ratio of Long to Short span,r = ly/l

x=

+ve Bending Moment Coefficient,αx+ =

-ve Bending Moment Coefficient,αx

- =

+ve Bending Moment Coefficient,αy+ =

-ve Bending Moment Coefficient, αy

- =

Factored +ve Bending Moment, Mux

+ = x 1.5w x lx

2 =

Factored -ve Bending Moment, Mux

- = x 1.5w x lx

2 =

Factored +ve Bending Moment, Muy

+ = x 1.5w x lx

2 =

Factored -ve Bending Moment, Muy

- = x 1.5w x lx

2 =

þÿInteriorPanel



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SUNIL HITECH ENGINEERS LIMITEDTCE-5910A-151-DC-

6857SHEL-PRL-720-CVC-

R-857RAW WATER PUMP HOUSE.DESIGN OF TWO WAY SLAB

AT EL(+)3.5 M

Reinforcement Calculations:

Short-span:

281.88

= 0.23 %

It is greater than minimum percentage of steel, 0.12%

= 1.20 %

Hence the section is Under reinforced section

Dia.Of bar used for +ve reinforcement= 10 mm

278.62 mm Spacing provided= 150 mm

say 150 mm

Whichever is lessMinimum spacing required = 3.d = 355.00 mm

300.00 mm

Provide Y10 @ 150 C/C as +ve reinforcement

Provided percentage of steel = 0.44 %

-ve reinforcement:

381.30

= 0.32 %


= 1.20 %


Dia.Of bar used for -ve reinforcement= 10 mm


say 150 mm


300.00 mm

Provide Y10 @ 150 C/C as -ve reinforcement


Long-span:

+ve reinforcement:

152.70

= 0.14 %


= 1.20 %


Dia.Of bar used for +ve reinforcement= 8 mm


say 200 mmWhichever is lessMinimum spacing required = 3.d = 330.00 mm

300.00 mm

Required Area of steel at midspan,Ast(+ve) =

Ast(reqd)

= mm2

Required percentage of steel, pt

= 100Ast/b.d

Limiting percentage of steel for U.R.S.,pt limit

= 19.94 x f ck

/f y

Spacing of the bars,Sv

= (b x Asv

) /Ast

Required Area of steel at support,Ast(-ve) =

Ast(reqd)

= mm2


= 100Ast/b.d


= 19.94 x f ck

/f y


= (b x Asv

) /Ast

Required Area of steel at midspan,Ast(+ve) =

Ast(reqd)

= mm2


= 100Ast/b.d


= 19.94 x f ck

/f y


= (b x Asv

) /Ast

f ck 2f y [

1−

1−

4 .598 M ubd 2 f ck ]

. bd

f ck

2f y [1− 1−4 .598

M u

bd 2 f ck ]. bd

f ck 2f

y [1−

1−

4 .598 M ubd 2 f ck ]

. bd



]




AT EL(+)3.5 M




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AT EL(+)3.5 M


-ve reinforcement:

205.25

= 0.18 %


= 1.20 %


Dia.Of bar used for -ve reinforcement= 8 mm Spacing provided= 200 mm

244.90 mm

say 200.00 mm


300.00 mm



Check for Deflection:

26

Percentage of tensile reinforcement = 0.44 %

1.65 (From IS:456)

1.00 (From IS:456)42.78

24.17 < (lx/d)max

Hence Provided depth is adequate.

Check for Shear:

28.64 kN

42.96 kN

0.36

0.44 %

0.46

This value is further enhanced by multiplying factor.

Multiplying factor ,K = 1.50

0.69 < Tv Hence OK

3.35 < Tv Hence OK

Hence Shear Strength is within the allowable limit.

FINAL RESULT:

Depth of the Slab = 150.00 mm

Reinforcement:

Short-span:


Provide Y10 @ 150 C/C as -ve reinforcementLong-span:


Required Area of steel at support,Ast(-ve) =

Ast(reqd)

= mm2


= 100Ast/b.d


= 19.94 x f ck

/f y


= (b x Asv

) /Ast

(lx/d

x) basic

=

Modification factor for Tension rein.K t=

Modification factor for Comp. rein.K c

=

(lx/d

x)

Max. = (l

x/d

x) basic

* K t* K

c=

(lx/d

x) provided

=

Maximum Shear Force, V = w x lx

/2 =

Factored Shear Force, Vu

=

Nominal Shear stress, Tv

= Vu/b.d = N/mm2

Percentage of tension reinforcement,pt=

Design Shear strength, Tc

= N/mm2

Therefore Design Shear Strength= KTc= N/mm2

Maximum shear strength,Tcmax

= N/mm2

f ck

2f y [1− 1−4 .598M u

bd 2 f ck ] . bd



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AT EL(+)3.5 M


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Two Way Slab at EL (+)3.5M