Grid Floors

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GRID FLOORS

INTRODUCTION

The floor which is resting on the beams running in two directions is known as

grid floor. In these types of floor, a mesh or grid of beams running in both the

directions is the main structure, and the floor is of nominal thickness. It is used to

cover a large area without obstruction of internal columns. They are generally

employed for architectural reasons for large rooms such as auditoriums, vestibules,

theatre halls, show rooms of shops where column free space is often the main

requirement.

An assembly of intersecting beams placed at regular interval and

interconnected to a slab of nominal thickness is known as Grid floor or Waffle floor.

These slabs are used to cover a large column free area and therefore are good choice

for public assembly halls. The structure is monolithic in nature and has more stiffness.

It gives pleasing appearance. The maintenance cost of these floors is less. However,

construction of the grid slabs is cost prohibitive.

By investigating various parameters the cost effective solution can be found for

the grid slabs, for which proper method of analysis need to be used. There are various

approaches available for analyzing the grid slab system. In present study some of

these approaches are studied and compared with each other. The comparison is done

on the basis of flexural parameters such as bending moments and shear forces

obtained from various methods. For carrying out study, halls having constant width

10.00m and varying ratio of hall dimensions (L/B) from 1 to 1.5 are considered.

Typical Plan of Grid System and Notations Used

Notations used:

1) L = Length of Hall (Longer side of hall)

2) B = Width of Hall (Shorter side of hall)

3) l = Spacing of grid beams in the direction of the length of the hall

4) b = Spacing of grid beams in the direction of the width of the hall

5) M

_x

= Bending moment in the beams running in x-direction

6) M

y

= Bending moment in the beams running in y-direction

7) Q

_x

= Shear force in the beams running in x-direction

8) Q

y

= Shear force in the beams running in y-direction

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Dimensions Considered

For the comparison purpose, the width of the hall is kept constant as 10.00 m

and length is increased by an interval of 1.00 m, so that L/B ratio varies from 1 to 1.5

at the interval of 0.1. For all hall sizes the thickness of grid slab is assumed as 100

mm. The size of the beams (0.23 m x 0.60 m) is kept same during entire study.

Analysis Of Grid Floors

(a) Approximate Methods

According to the Indian Standard Code IS: 456-1978, the ribbed slab system can be

analysed as solid slab, if the following requirements regarding spacing of beams,

thickness of slab and edge beams are satisfied.

1. l. The spacing Of the ribs should not be greater than 1.5 m and it should not be

greater than 12 times the flange thickness.

2. In situ ribs should not be less than 65 mm wide.

3. The ribs should be formed along each edge parallel to the span, having a width

equal to that of the bearing.

The moments and shears per unit width Of grid are determined from Table 22 Of IS :

456 code and the reinforcements are designed in the ribs.

The slab reinforcement generally consists of a mesh or fabric. A second approximate

method which is applicable to the grid floor system is the Rankine Grashoff theory of

equating deflections at the junctions of ribs.

Dimensions Considered

For the comparison purpose, the width of the hall is kept constant as 10.00 m

and length is increased by an interval of 1.00 m, so that L/B ratio varies from 1 to 1.5

at the interval of 0.1. For all hall sizes the thickness of grid slab is assumed as 100

mm. The size of the beams (0.23 m x 0.60 m) is kept same during entire study.

Analysis Of Grid Floors

(a) Approximate Methods

According to the Indian Standard Code IS: 456-1978, the ribbed slab system can be

analysed as solid slab, if the following requirements regarding spacing of beams,

thickness of slab and edge beams are satisfied.

1. l. The spacing Of the ribs should not be greater than 1.5 m and it should not be

greater than 12 times the flange thickness.

2. In situ ribs should not be less than 65 mm wide.

3. The ribs should be formed along each edge parallel to the span, having a width

equal to that of the bearing.

The moments and shears per unit width Of grid are determined from Table 22 Of IS :

456 code and the reinforcements are designed in the ribs.

The slab reinforcement generally consists of a mesh or fabric. A second approximate

method which is applicable to the grid floor system is the Rankine Grashoff theory of

equating deflections at the junctions of ribs.

Dimensions Considered

For the comparison purpose, the width of the hall is kept constant as 10.00 m

and length is increased by an interval of 1.00 m, so that L/B ratio varies from 1 to 1.5

at the interval of 0.1. For all hall sizes the thickness of grid slab is assumed as 100

mm. The size of the beams (0.23 m x 0.60 m) is kept same during entire study.

Analysis Of Grid Floors

(a) Approximate Methods

According to the Indian Standard Code IS: 456-1978, the ribbed slab system can be

analysed as solid slab, if the following requirements regarding spacing of beams,

thickness of slab and edge beams are satisfied.

1. l. The spacing Of the ribs should not be greater than 1.5 m and it should not be

greater than 12 times the flange thickness.

2. In situ ribs should not be less than 65 mm wide.

3. The ribs should be formed along each edge parallel to the span, having a width

equal to that of the bearing.

The moments and shears per unit width Of grid are determined from Table 22 Of IS :

456 code and the reinforcements are designed in the ribs.

The slab reinforcement generally consists of a mesh or fabric. A second approximate

method which is applicable to the grid floor system is the Rankine Grashoff theory of

equating deflections at the junctions of ribs.

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Consider a grid floor shown in Fig. in Which the spacings of the ribs are a 1 and b1 in

the x and y directions respectively.

q = Total load per unit area and q1 and q2 are the loads shared in the x and y

directions

a= Shorter dimension of grid

b= Longer dimension of grid

The deflections of the ribs AB and CD at the junction O must be the same and by

equating the deflections,

we have deflection =

5

384 =

5

384 =

=

+

solving the above equations we have

=

and

=

Consider a grid floor shown in Fig. in Which the spacings of the ribs are a 1 and b1 in

the x and y directions respectively.

q = Total load per unit area and q1 and q2 are the loads shared in the x and y

directions

a= Shorter dimension of grid

b= Longer dimension of grid

The deflections of the ribs AB and CD at the junction O must be the same and by

equating the deflections,

we have deflection =

5

384 =

5

384 =

=

+

solving the above equations we have

=

and

=

Consider a grid floor shown in Fig. in Which the spacings of the ribs are a 1 and b1 in

the x and y directions respectively.

q = Total load per unit area and q1 and q2 are the loads shared in the x and y

directions

a= Shorter dimension of grid

b= Longer dimension of grid

The deflections of the ribs AB and CD at the junction O must be the same and by

equating the deflections,

we have deflection =

5

384 =

5

384 =

=

+

solving the above equations we have

=

and

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and bending moments for the ribs are given by

=

₈

=

₈

The bending moments in the other ribs can also be determined in direct proportion to

their distances from the centre. The ribs are designed as flanged sections to resist the

moments and shears. However the approximate methods do not yield the twisting

moments in the beams. For small span grids with spacings of ribs not exceeding 1.5

metres, approximate methods can be used, but for grids of larger spans With spacing s

of ribs exceeding I .5 m, a rigorous analysis based on orthotropic plate theory is

generally used.

Approximate Analysis Of Grid Floors

Floors with restricted layout of beams, thickness of slab and edge beams can be

analysed by the conventional methods as in ribbed slabs. Large grid floors which do

not follow these restricted layouts are analysed by other methods. These methods can

be divided into three groups:

1. Method based on plate theory (approximate method)

2. Stiffness matrix method using computer

3. Equating deflection method of each intersecting node oy simultaneous

equations.

In the following sections, a brief discussion of the method based on the plate theory

and reference to other two methods

Analysis Of Rectangular Grid Floors By Timoshenko's Plate Theory

Many variations in the plate theory by different authors are available for analysis of

grid floors. The most popular one is that given by Timoshenko and Krieger t I J. They

have shown that the moments and shears in an anisotropic plate, freely supported on

four sides, depend on the deflection surface. The vertical deflection (D at any point of

a symmetrical grid shown in Fig

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Analysis Of Grid By Stiffness Matrix Method

This is an exact method and needs a computer for its application. It assumes the frame

coordinates system X and Y and the member coordinate system X'—Y' as shown in

Fig. 6.1(b). The active joints are identified first and then the force displacement

equations are written for the various active member ends. The member stiffnesses are

then transferred to the frame coordinate system. From the stiffness matrix and joint

displacement equations, the force at each member joint can be obtained by use of

computer programs. Readymade programs for the various types of grids are also

available for quick analysis in design offices. If a computer and the relevant software

programs are available, the analysis can be carried out quickly. However, the results

of the computer analysis should be checked with values obtained from approximate

methods.

Analysis Of Grid Floors By Equating Joint Deflections

In this method, the deflection of the joints and the loads on each Of the beams are

determined. As the number of resulting equations is large, a computer will be required

for its easy solution. (Analysis of Grid Floors ) by the National Buildings

Organisation, New Delhi gives tables and charts which may be conveniently used for

the analysis of rectangular and diagonal grid floors• In this method, torsion component

is generally neglected

Analysis Of Grid By Stiffness Matrix Method

This is an exact method and needs a computer for its application. It assumes the frame

coordinates system X and Y and the member coordinate system X'—Y' as shown in

Fig. 6.1(b). The active joints are identified first and then the force displacement

equations are written for the various active member ends. The member stiffnesses are

then transferred to the frame coordinate system. From the stiffness matrix and joint

displacement equations, the force at each member joint can be obtained by use of

computer programs. Readymade programs for the various types of grids are also

available for quick analysis in design offices. If a computer and the relevant software

programs are available, the analysis can be carried out quickly. However, the results

of the computer analysis should be checked with values obtained from approximate

methods.

Analysis Of Grid Floors By Equating Joint Deflections

In this method, the deflection of the joints and the loads on each Of the beams are

determined. As the number of resulting equations is large, a computer will be required

for its easy solution. (Analysis of Grid Floors ) by the National Buildings

Organisation, New Delhi gives tables and charts which may be conveniently used for

the analysis of rectangular and diagonal grid floors• In this method, torsion component

is generally neglected

Analysis Of Grid By Stiffness Matrix Method

This is an exact method and needs a computer for its application. It assumes the frame

coordinates system X and Y and the member coordinate system X'—Y' as shown in

Fig. 6.1(b). The active joints are identified first and then the force displacement

equations are written for the various active member ends. The member stiffnesses are

then transferred to the frame coordinate system. From the stiffness matrix and joint

displacement equations, the force at each member joint can be obtained by use of

computer programs. Readymade programs for the various types of grids are also

available for quick analysis in design offices. If a computer and the relevant software

programs are available, the analysis can be carried out quickly. However, the results

of the computer analysis should be checked with values obtained from approximate

methods.

Analysis Of Grid Floors By Equating Joint Deflections

In this method, the deflection of the joints and the loads on each Of the beams are

determined. As the number of resulting equations is large, a computer will be required

for its easy solution. (Analysis of Grid Floors ) by the National Buildings

Organisation, New Delhi gives tables and charts which may be conveniently used for

the analysis of rectangular and diagonal grid floors• In this method, torsion component

is generally neglected

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Comparison Of Methods Of Analysis

No torsion analysis is suitable for the preliminary analysis and design. The torsion

analysis method gives fairly good results for rectangular grids. The handbook method

gives best results for square grids. However, complex grids, like diagonal grids, are

best solved by matrix method for final designs. Even though the slab analogy of

voided slabs is straightforward, its use should be restricted. For this the following

conditions are necessary:

1. Spacing of ribs should not be greater than 1.5 m or twelve times the flange

thickness (the clear spacing should not be more than ten times the thickness).

2. The depth of the rib should not be more than four times the width of the ribs•

3. With large spacing of the beams, grids do not act as a plate but as individual

units.

Accordingly' with the above spacing restrictions the plate analogy can be used for

approximate analysis Of rectangular grids. For grids not obeying the above spacing

rules, the approximate methods can be used for the preliminary analysis only. In all

the cases, the matrix method gives more realistic values and should be used for the

final analysis and design.

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A reinforced concrete grid floor of size 9m x 9m is required for an assembly

hall. Assuming rib spacing of 1.5 m in the span of short direction and 1.5m in

the long span direction, design the grid floor. Adopt M20 grade concrete and

Fe415 grade steel. Live load may be assumed as 4KN/m

2

Solution:

1)Dimension of the Slab:

Adopt thickness of slab =100mm Depth of Ribs:

Span/Depth = 20 (Simply supported case) Depth = (9 x 103)/20 = 450 mm Width of rib =200mm

Adopt overall depth of ribs =550mm 2)Loads:

Weight of the slab =(0.1x24) =2.4KN/m2 Total load of slab =(2.4x9x9) =194.4KN Weight of ribs =(.2x.45x24) =2.16KN/m Total weight of beam(Y-direction) =(7x2.16x9) =136.08KN Total weight of beam(X-direction) =(7x2.16x9) =136.08KN Total weight of Floor Finish =(0.6x9x9) =48.6KN Total live load =(4x9x9) =324KN Total (Dead load+Live Load) =843.72 KN

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3) Analysis:

a) Section Properties : = = 0.181

= = 0.133 (X- direction) = = 0.133 ( Y-direction) From design table

I = CbwD3

= (0.191x200x5503) (C=0.191) = 63.8x108mm4= I1=I2

Where I1and I2are the second moment of inertia of T-sections about their centrodial axis in the

x and y directions respectively. D1 = EI1 and D2= EI2

Dx= (EI1/b1) =(E x 63.8x108/1.5x1012) =0.00425E

Dy= (EI1/a1) =(E x 63.8x108/1.5x1012) =0.00425E

b) The Torsional Rigidity in X and Y directions are given by Cx=GC1/b1 and Cy= GC2/a1 Assume µ=0.15 ; E= 5700 = 5700 =25.49 x 106KN/m2 G = = 0.435µ G = 11.09x106KN/m2 C1= C2 = (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/055))( ) C1=C2= 1.13 X10-3

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Cx= GC1/b1 = = 8.35x103 Cy= GC2/a1 = = 8.35x103 2H = Cx +Cy = 16.7 x 103m unit

3) Central Deflection: (a= 9m , b=9m)

= = 12.34 --- (A) = = 12.34 --- (B)

= = 2.545 ---(C) A+B+C = 27.225

Central Deflection (W1) = = = 8.82 m

Long term deflection is equal to 2 to 3 times elastic deflection depending on the date of removal of supports.

Long term deflection = 3 x 8.82 = 26.46 mm

Span /250 = 9000/250 = 36 >26.46 (IS 456-2000) Maximum deflection is within the permissible limits.

4) Design Moments and Shear:

Mx = -(Dx ) = ( )2 [ ] =0.85x105( )2x0.0088

= 117.46

My = -(Dy ) = Dy ( )2 [ ] =1.083x105( )2x0.0088

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= 115.736 Mxy = (-)( ) ( )2 [ ] = (-) ( )( )x0.0088x = (-) 6.53 Myx = (-) 6.05 Qx = (Dx + ) = (-) ( )(Dx + Dy ) = (-) 0.0088((0.85x105x ) +8.35x103x )) ( ) = -37.02( ) Qy = (-) {(Dx( )3+Cx( )}w1{( ) =(-) {(0.85x105x ( )3+ 6.265x103( )}x 0.008( ) Qy = - 34.749 ( ) Points x,y A 4.5,6 1 0 1 0 B 3,6 0.866 0.5 1 0 C 1.5,6 0.5 0.866 1 0 D 0,6 0 0 1 1 0 E 0,10 0 0 1 0.5 -0.866 F 0,12 0 0 1 0 -1 G 4.5,12 1 0 0 -1 H 4.5,10 1 0 0.5 -0.866 Table-4.1

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POINTS MOMENTS(KNm) SHEAR(KN) Mx My Mxy Myx Qx Qy A 117.46 115.736 0 00 0 0 B 105.401 107.361 0 0 16.76 0 C 72.075 74.031 0 0 29.082 0 D 0 0 0 0 37.02 0 E 0 0 5.95 6.05 16.76 0 F 0 0 5.53 6.05 0 0 G 0 0 0 0 0 34.749 H 56.745 57.25 0 0 0 22.13 Table-4.2 5) Design of Reinforcement: Max moment = 117.46 KNm

Moment resisted by central rib in x-direction over 1.5m width = 1.5x117.46 = 176.5735 KNm Ultimate moment = 264.428 KNm

Moment Capacity of Flange section Muf= 0.36fckbfDf(d-0.42Df)

= ( 0.36x20x1.5x103x100)(500-0.42x100) = 494.64 KNm > 264.86 KNm

Since Mu < Muf Neutral axis falls within the flange Mu = 0.87fyAstd (1- )

100 (Ast/bd) = pt = (1- ) = (1- )

= 1.1629x 10-3 1162.93 mm2

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No of bars = = 4

Provide 4 bars of 20 mm diameter (Ast= 1256 mm2)

Shear Reinforcement:

Maximum ultimate shear = (37.02) x 1.5 x 1.5 = 83.295 KN

v = = = 0.932 N/mm2

Assuming two bars to b bent up at the supports, Astat supports = 628 mm2

Pt = = = 0.628

c = 0.528 N/mm2 < v

Vud = = (1.0056-0.5218) x 200 x 500

= 48.432x103 = 0.2684

Provide 8 mm dia bars 2 legged , = 2x50.24 mm2 = 0.2648

= 374.255 3d =3x500 = 1500m

Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Y- Direction: M =115.736KNm Mu = 2x115.736x1.5 =326.205 KNm R = = = 0.655 (Ast/bd) = pt = (1- ) Ast =1062.1125 mm2

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For 20 mm diameter bars, no of bars required = 4

st= 1256 mm2)

For shear reinforcement provid 8 mm diameter bars @ 400 mm c/c SLAB:

Mxy = 6.0 KNm Myx = 6.05 KNm

Since the moments being small in the slab, mesh reinforcement consisting of 6 mm diameter at 200 mm c/c provided both ways for positive and negative Bending Moment.

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A reinforced concrete grid floor of size 10m x 14m is required for an assembly

hall. Assuming rib spacing of 2 m in the span of short direction and 2m in the

long span direction, design the grid floor. Adopt M20 grade concrete and

Fe415 grade steel. Live load may be assumed as 4KN/m

2

Solution:

Span/Depth = 20 (Simply supported case) Depth = (10x103)/20 = 500mm Width of rib =200mm

Weight of the slab = (0.1x24) =2.4KN/m2 Total load of slab = (2.4x10x14) =336KN Weight of ribs = (.2x.6x24) =2.88KN/m Total weight of beam(Y-direction) = (8x2.88x14) =322.56KN Total weight of beam(X-direction) = (6x2.88x10) =172.8KN Total weight of Floor Finish = (0.6x10x14) =84KN Total live load = (4x10x14) =560KN Total (Dead load+Live Load) = 1475.36KN

Load per meter =1475.36/(10x14) = 10.53KN/m2 3) Analysis:

a) Section Properties : Df/Dw=100/600 = 0.166

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bw/bf=200/2000 = 0.1 (X- direction)

=200/2000 = 0.1 ( Y-direction) From design table

I = CbwD3

= (0.191x200x6003) (C=0.191) = 82.5x108mm4= I1=I2

Dx= (EI1/b1) =(E x 82.5x108/2x1012) =0.00319E

Dy= (EI1/a1) =(E x 82.5x108/2x1012) =0.00319E

b) The Torsional Rigidity in X and Y directions are given by Cx=GC1/b1 and Cy= GC2/a1 Assume µ=0.15 ; E= 5700 = 5700 =25.49 x 106KN/m2 G= (E/2(1+µ)) = 0.435µ G= 11.09x106KN/m2 C1=C2= (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/0.55))((0.23x0.55)/3) C1=C2= 1.13 X10-3 Cx= GC1/b1 = (((11.09x106)(1.13x10-3))/2)= 6.26x103 Cy= GC2/a1 = (((11.09x106)(1.13x10-3))/2)= 6.26x103 2H = Cx +Cy = 12.52 x 103m unit

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3) Central Deflection: (a= 10m , b=14m) (Dx/a4) = ((105x0.81)/104) = 8.1 --- (A) (Dy/b4) = ((105x0.81)/144) =2.10 --- (B) ((2H)/(a2b2)) =((12.52x103)/(102142))= 0.638 ---(C) A+B+C = 10.83 Central Deflection (W1) = ((16q)/(10.83 6)) = ((16x10.53)/(10.83 6)) = 16.2 mm

Mx = -(Dx ) = Dx( )2 [ ] = 0.81x105( )2x0.0194 = 154.93 My = (-) (Dy ) = (-) Dy ( )2 [ ] = (-) 0.81x105( )2x0.0194 = (-) 79.04 Mxy = (-)( ) = -( ) ( )2 [ ] = (-) ( )( )2x0.0194x

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= -5.51 Myx = -5.51 Qx = (Dx + ) = (-) ( )(Dx + Dy ) = (-) 0.0194((0.81x105x ) +48.02x )) ( ) = (-)154.93( ) Qy = (-) {(Dx( )3+Cx( )}w1{( ) = (-) {(0.81x105x ( )3+ 6.26x103( )}x 0.0194( ) Qy = - 18.58 ( ) Points x,y A 5,7 1 0 1 0 B 4,7 0.021 1 1 0 C 2,7 0.01 1 1 0 D 0,7 0 0 1 1 0 E 0,12 0 0 1 0.04 1 F 0,14 0 0 1 0 -1 G 5,12 1 0 0.04 1 H 5,14 1 0 0 -1 Table-4.1

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POINTS MOMENTS(KNm) SHEAR(KN) Mx My Mxy Myx Qx Qy A 154.93 79.04 0 00 0 0 B 3.25 1.65 0 0 -154.93 0 C 1.54 0.79 0 0 -154.93 0 D 0 0 0 0 -154.93 0 E 0 0 -5.51 -5.51 -6.19 0 F 0 0 5.51 5.51 0 0 G 0 0 0 0 -6.19 -18.58 H 0 0 0 0 0 18.58 Table-4.2 5) Design of Reinforcement: Max moment = 154.93 KNm

Moment resisted by central rib in x-direction over 2m width = 2x154.93 = 309.86 KNm

Ultimate moment = 464.79 KNm

= (0.36x20x2x103x100)(550-0.42x100) = 731.52 KNm > 464.79 KNm

100 (Ast/bd) = pt = (1- ) = (1- )

= 1.1629x 10-3 1162.93 mm2

Choose 20 diameter bars, Astb= 314 mm2

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Maximum ultimate shear = (154.93) x 2 x 1.5 = 464.79 KN

v = = = 4.22 N/mm2

Pt = = = 0.9

c = 0.599 N/mm2 < v

Vud = = (4.22-0.599) x 200 x 550

= 398.31Sx103 = 1.96

Provide 10 mm dia bars 2 legged , = 2x157 mm2 = 1.96

= 160

3d =3x550 = 1650m

Provide 2 legged 10 mm diameters bars @ 200mm c/c. Mu Along Y- Direction: M = 79.4 KNm Mu = 2x79.4x1.5 = 238.2 KNm R = = = 3.93 (Ast/bd) = pt = (1- ) Ast =1827.23 mm2

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Provide 25 st= 1960 mm2)

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A reinforced concrete grid floor of size 9m x 12m is required for an assembly

hall. Assuming rib spacing of 1.5 m in the span of short direction and 2m in

the long span direction, design the grid floor. Adopt M20 grade concrete and

Fe415 grade steel. Live load may be assumed as 4KN/m

2

Solution:

Weight of the slab =(0.1x24) =2.4KN/m2 Total load of slab =(2.4x9x12) =259.24KN Weight of ribs =(.2x.45x24) =2.16KN/m Total weight of beam(Y-direction) =(7x2.16x12) =181.44KN Total weight of beam(X-direction) =(7x2.16x9) =136.08KN Total weight of Floor Finish =(0.6x9x12) =64.8KN Total live load =(4x9x12) =432KN Total (Dead load+Live Load) =1073.2 KN

Load per meter = =9.94KN/m2 3) Analysis:

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= = 0.181

I = CbwD3

= (0.191x200x5503) (C=0.191) = 63.8x108mm4= I1=I2

Dx= (EI1/b1) =(E x 63.8x108/2x1012) =0.00319E

Dy= (EI1/a1) =(E x 63.8x108/1.5x1012) =0.00425E

b) The Torsional Rigidity in X and Y directions are given by Cx=GC1/b1 and Cy= GC2/a1 Assume µ=0.15 ; E= 5700 = 5700 = 25.49 x 106KN/m2 G= = 0.435µ G= 11.09x106KN/m2 C1=C2= (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/055))( ) C1=C2= 1.13 X10-3 Cx= GC1/b1 = = 6.26x103 Cy= GC2/a1 = = 8.35x103

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2H = Cx +Cy = 14.61 x 103m unit 3) Central Deflection: (a= 9m , b=12m)

= = 12.34 --- (A) = = 5.22 --- (B) = = 1.225 ---(C) A+B+C = 18.81

Central Deflection (W1) = = = 8.82 mm

Mx = -(Dx ) = ( )2 [ ] = 0.85x105( )2x0.0088 = 91.049 My = -(Dy ) = Dy ( )2 [ ] = 1.083x105( )2x0.0088 = 65.25 Mxy = -( ) = -( ) ( )2 [ ]

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=(-) ( )( )x0.0088x =(-) 4.53 Myx = -6.05 Qx = (Dx + ) =(-) ( )(Dx + Dy ) =(-)0.0088((0.85x105x ) +8.35x103x )) ( ) = -33.52( ) Qy = (-){(Dx( )3+Cx( )}w1{( ) = (-){(0.85x105x ( )3+ 6.265x103( )}x 0.008( ) Qy = - 15.57 ( ) Points x,y A 4.5,6 1 0 1 0 B 3,6 0.866 0.5 1 0 C 1.5,6 0.5 0.866 1 0 D 0,6 0 0 1 1 0 E 0,10 0 0 1 0.5 -0.866 F 0,12 0 0 1 0 -1 G 4.5,12 1 0 0 -1 H 4.5,10 1 0 0.5 -0.866 Table-4.1

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Moment resisted by central rib in x-direction over 1.5m width = 1.5x91.049 = 136.5735 KNm

= ( 0.36x20x1.5x103x100)(500-0.42x100) = 494.64 KNm > 204.86 KNm

100 (Ast/bd) = pt = (1- ) = (1- )

= 1.1629x 10-3 1162.93 mm2

Choose 20 diameter bars, Astb= 314 mm2

(32)

Maximum ultimate shear = (33.52) x 2 x 1.5 = 100.56 KN

v = = = 1.0056 N/mm2

Pt = = = 0.628

c = 0.528 N/mm2 < v

Vud = = (1.0056-0.5218) x 200 x 500

= 48.432x103 = 0.2684

= 374.255 3d =3x500 = 1500m

Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Y- Direction: M = 62.5 KNm Mu = 2x62.5x1.5 = 187.5 KNm R = = = 0.375 (Ast/bd) = pt = (1- ) Ast =1062.1125 mm2

(33)

Provide 20 mm diameter bars of 4 st= 1256 mm2)

(34)

(35)

(36)

A reinforced concrete grid floor of size 10m x 12m is required for an assembly

hall. Assuming rib spacing of 2 m in the span of short direction and 2m in the

long span direction, design the grid floor. Adopt M20 grade concrete and

Fe415 grade steel. Live load may be assumed as 4KN/m

2

Solution:

Weight of the slab =(0.1x24) =2.4KN/m2 Total load of slab =(2.4x9x12) =288KN Weight of ribs =(.2x.45x24) =2.16KN/m Total weight of beam(Y-direction) =(7x2.16x12) =120KN Total weight of beam(X-direction) =(7x2.16x9) =172KN Total weight of Floor Finish =(0.6x9x12) =72KN Total live load =(4x9x12) =480KN Total (Dead load+Live Load) =1132.8 KN

(37)

3) Analysis:

I = CbwD3

= (0.191x200x6003) (C=0.191) = 82.51x108mm4= I1=I2

Dx= (EI1/b1) =(E x 82.51x108/2x1012) =0.00412E

Dy= (EI1/a1) =(E x 82.51x108/2x1012) =0.00412E

b) The Torsional Rigidity in X and Y directions are given by Cx=GC1/b1 and Cy= GC2/a1 Assume µ=0.15 ; E= 5700 = 5700 =25.49 x 106KN/m2 G= = 0.435µ G= 11.09x106KN/m2 C1=C2= (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/.6))( ) C1=C2= 1.264 X10-3

(38)

Cx= GC1/b1 = = 7.002x103 Cy= GC2/a1 = = 7.002x103 2H = Cx +Cy = 14.001 x 103m unit

= = 10.5 --- (A) = = 5.06 --- (B)

= = .9722 ---(C) A+B+C = 16.53

Central Deflection (W1) = = = 9.3mm

Mx = -(Dx ) = ( )2 [ ] = 0.85x105( )2x0.0093

= 116.24

My = -(Dy ) = Dy ( )2 [ ] =1.083x105( )2x0.0093

(39)

= 67.25 Mxy = -( ) = -( ) ( )2 [ ] = (-) ( )( )x0.0088x = -4.73 Myx = -6.85 Qx = (Dx + ) = (-) ( )(Dx + Dy ) = (-) 0.00938((0.85x105x ) +8.35x103x )) ( ) = (-) 36.122( ) Qy = (-){(Dx( )3+Cx( )}w1{( ) = (-) {(0.85x105x ( )3+ 6.265x103( )}x 0.008( ) Qy =(-) 15.57 ( ) Points x,y A 4.5,6 1 0 1 0 B 3,6 0.866 0.5 1 0 C 1.5,6 0.5 0.866 1 0 D 0,6 0 0 1 1 0 E 0,10 0 0 1 0.5 -0.866 F 0,12 0 0 1 0 -1 G 4.5,12 1 0 0 -1 H 4.5,10 1 0 0.5 -0.866 Table-4.1

(40)

Moment resisted by central rib in x-direction over 2m width = 2x116.24 = 174.24 KNm

= ( 0.36x20x2x103x100)(550-0.42x100) = 548.64 KNm > 261.54 KNm

100 (Ast/bd) = pt = (1- ) = (1- )

= 1.39x 10-3 1396..21 mm2

Choose 25 diameter bars, Astb= 785mm2

(41)

Maximum ultimate shear = (36.52) x 2 x 2 = 138.08 KN

v = = = 1.218 N/mm2

Pt = = = 0.628

c = 0.528 N/mm2 < v

Vud = = (1.218-0.5218) x 200 x 550

= 48.432x103 = 0.2684

= 262.91

3d =3x550 = 1650m

Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Y- Direction: M = 67.5 KNm Mu = 2x67.5x2= 270 KNm R = = = 0.462 (Ast/bd) = pt = (1- ) Ast =1440.71 mm2

(42)

Provide 20 mm st= 1519mm2)

Mxy = 4.73 KNm Myx = 6.88KNm

(43)

(44)

(45)

**A reinforced concrete grid floor of size 12m*16 is required for an assembly**

hall. Assuming rib spacing of 2 m in the span of short direction and 2m in the

long span direction, design the grid floor. Adopt M20 grade concrete and

Fe415 grade steel. Total load including self weight is 6.5 KN/m^2

Solution:

Estimate the necessary thickness of the beam and floor slab Assume the spacing of ribs 2 m both ways

Lx= 12m, Df =12000/20=600mm Lx=2m, Df=2000/20=100mm Assume width of rib=200mm 2)Loads:

Load per meter =6.5KN/m2 3) Analysis:

= = 10(X- direction) = = 0.133 ( Y-direction) From design table

I = CbwD3

= (0.191x200x6003) (C=0.191) = 82.6*10^-4M^4 = I1=I2

(46)

Dx= (EI1/b1) =(25.49*10^6 x 82.8x108/2x1012) =1.055*10^5M

Dy= (EI1/a1) =(25.49*10^6x 82.8x108/2x1012) =1.055*10^5M

b) The Torsional Rigidity in X and Y directions are given by Cx=GC1/b1 and Cy= GC2/a1 Assume µ=0.15 ; E= 5700 = 5700 =25.49 x 106KN/m2 G = = 0.435µ G= 11.09x106KN/m2 C1=C2= (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/0.6))( ) C1=C2= 1.264 X10-3 Cx= GC1/b1 = = 7x103 Cy= GC2/a1 = = 7x103 2H = Cx +Cy = 14 x 103m unit

= = 5.09 --- (A) = = 1.61 --- (B)

= = .38 ---(C) A+B+C = 7.08

(47)

Long term deflection = 2 x 15 = 30 mm

Span /250 = 12000/(2.5*15) = 320 <250(allowed) (IS 456-2000) Maximum deflection is within the permissible limits.

Mx = -(Dx ) = ( )2 [ ] =0.85x105( )2x0.0088 = 91.049 My = -(Dy ) = Dy ( )2 [ ] =1.083x105( )2x0.0088 = 65.25 Mxy = -( ) = -( ) ( )2 [ ] = (-)( )( )x0.0088x = (-)4.53 Myx = -6.05 Qx = (Dx + ) = (-) ( )(Dx + Dy )

(48)

= -0.0088((0.85x105x ) +8.35x103x )) ( ) = -33.52( ) Qy = -{(Dx( )3+Cx( )}w1{( ) =-{(0.85x105x ( )3+ 6.265x103( )}x 0.008( ) Qy = - 15.57 ( ) Points x,y A 6 8 1 0 1 0 B 4,8 0.866 0.5 1 0 C 2.8, 0.5 0.866 1 0 D 0,8 0 0 1 1 0 E 0,12 0 0 1 0.5 -0.866 F 0,16 0 0 1 0 -1 G 6,16 1 0 0 -1 H 0.5,10 1 0 0.5 -0.866 Table-4.1 POINTS MOMENS (KNm) SHEAR (KN) POINTS MOMENTS (KNm) SHEAR (KN) POINTS Mx My Mx My A 108 61 A 108 61 A B 94 53 B 94 53 B C 54 30.5 C 54 30.5 C D 0 0 D 0 0 D E 0 0 E 0 0 E F 0 0 F 0 0 F G 0 0 G 0 0 G H 77 43 H 77 43 H Table-4.2 5) Design of Reinforcement: Max moment = 108 KNm

Moment resisted by central rib in x-direction over 2m width = 2x108 = 216 KNm

(49)

=( 0.36x20x2x103x100)(500-0.42x100) = 659 KNm > 324KNm

100 (Ast/bd) = pt = (1- ) = (1- )

= 2078x 10-3 2078 mm2

Choose 25 diameter bars, Astb= 490.8mm2

No of bars = = 4.23

Provide 4 bars of 25 mm and 1 bar of 20 mm diameter (Ast= 2277 mm2)

Maximum ultimate shear = (29.52) x 2 x 2 = 118.08 KN

v = = = .984 N/mm2

Pt = = = 0.523

c = 0.523 N/mm2 < v

Vud = = (.984-0.523) x 200 x 600

= 55.32x103 = 0.255

(50)

= 0.255 = 394.039 3d =3x500 = 1500m

Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Y- Direction: M = 61KNm Mu = 2x61x1.5 = 183.5 KNm R = = = 2.54 (Ast/bd) = pt = (1- ) Ast =1030.1125 mm2

For 20 mm diameter bars, no of bars required = 4

st= 1256 mm2)

(51)

(52)