Grid Floors

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) Mx= Bending moment in the beams running in x-direction 6) My= Bending moment in the beams running in y-direction 7) Qx= Shear force in the beams running in x-direction 8) Qy= Shear force in the beams running in y-direction

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.

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

solving the above equations we have =

=

and

=

=

=

5 384

+

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

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

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.

EXAMPLES

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/m2 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

Load per meter

=

=10.416KN/m2

3) Analysis: a) Section Properties : =

= 0.181

=

= 0.133 (X- direction)

=

= 0.133 ( Y-direction)

From design table = CbwD3

I

= (0.191x200x5503)

(C=0.191)

= 63.8x108mm4 = I1 =I2 Where I1 and I2 are 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

=25.49 x 106 KN/m2 G=

= 0.435µ

G = 11.09x106 KN/m2 C1= C2 = (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/055))( C1=C2= 1.13 X10-3

)

= 5700

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

= 36 >26.46

= 9000/250

(IS 456-2000)

Maximum deflection is within the permissible limits. 4) Design Moments and Shear: Mx = -(Dx )

=

( )2

[

]

=0.85x105( )2 x0.0088 = 117.46 My = -(Dy )

= Dy ( )2

[

=1.083x105( )2 x0.0088

]

= 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(

= - 34.749 (

Points

x,y

A B C D E F G H

4.5,6 3,6 1.5,6 0,6 0,10 0,12 4.5,12 4.5,10

)

) )}w1{(

)

=(-) {(0.85x105x ( )3 + 6.265x103( Qy

)) (

)}x 0.008(

)

)

0 0 0

1 0 0.866 0.5 0.5 0.866 0 1 0 1 0 1 1 0 1 0 Table-4.1

1 1 1 1 0.5 0 0 0.5

0 0 0 0 -0.866 -1 -1 -0.866

POINTS

MOMENTS(KNm) My Mxy 115.736 0 107.361 0 74.031 0 0 0 0 5.95 0 5.53 0 0 57.25 0 Table-4.2

Mx 117.46 105.401 72.075 0 0 0 0 56.745

A B C D E F G H

Myx 00 0 0 0 6.05 6.05 0 0

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 Choose 20 diameter bars, Astb = 314 mm2

SHEAR(KN) Qx 0 16.76 29.082 37.02 16.76 0 0 0

Qy 0 0 0 0 0 0 34.749 22.13

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, Ast at supports = 628 mm2 Pt = c

=

= 0.628

= 0.528 N/mm2 <

Vud

=

v

= (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=

=

(Ast/bd) = pt = Ast

= 0.655 (1-

= 1062.1125

mm2

)

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.

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/m2 Solution: 1)Dimension of the Slab: Adopt thickness of slab

=100mm

Depth of Ribs: Span/Depth

= 20 (Simply supported case)

Depth

= (10x103)/20 = 500mm

Width of rib

=200mm

Adopt overall depth of ribs

=600mm

2)Loads: =2.4KN/m2

Weight of the slab

= (0.1x24)

Total load of slab

= (2.4x10x14) =336KN

Weight of ribs

= (.2x.6x24)

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)

=2.88KN/m

=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

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 Where I1 and I2 are 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 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

Assume µ=0.15

; E= 5700

Cy= GC2/a1 = 5700

=25.49 x 106 KN/m2 G= (E/2(1+ µ)) = 0.435µ G= 11.09x106 KN/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

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

Long term deflection is equal to 2 to 3 times elastic deflection depending on the date of removal of supports. Long term deflection = 2 x 16.2 Span /250

= 32.4 mm

= 10000/250

= 40 >32.4

(IS 456-2000)

Maximum deflection is within the permissible limits. 4) Design Moments and Shear: = Dx( )2

Mx = -(Dx )

[

]

= 0.81x105( )2 x0.0194 = 154.93 My = (-) (Dy )

= (-) Dy ( )2

[

]

= (-) 0.81x105( )2 x0.0194 = (-) 79.04 Mxy = (-)(

)

= -(

= (-) (

) ( )2

[

]

)(

)2x0.0194x

= -5.51 Myx

= -5.51

Qx

= = (-)

Qy

(Dx

+

)

(

)(Dx

+ Dy

= (-) 0.0194((0.81x105x

) +48.02x

= (-)154.93(

)

= (-) {(Dx( )3 +Cx(

)}w1{(

= - 18.58 (

Points

x,y

A B C D E F G H

5,7 4,7 2,7 0,7 0,12 0,14 5,12 5,14

)) (

)

)

= (-) {(0.81x105x ( )3 + 6.26x103( Qy

)

)}x 0.0194(

)

)

0 0 0

1 0.021 0.01 0 0 0 1 1

0 1 1 1 1 1 0 0 Table-4.1

1 1 1 1 0.04 0 0.04 0

0 0 0 0 1 -1 1 -1

POINTS

MOMENTS(KNm) My Mxy 79.04 0 1.65 0 0.79 0 0 0 0 -5.51 0 5.51 0 0 0 0 Table-4.2

Mx 154.93 3.25 1.54 0 0 0 0 0

A B C D E F G H

Myx 00 0 0 0 -5.51 5.51 0 0

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

Moment Capacity of Flange section Muf= 0.36fckbfDf(d-0.42Df) = (0.36x20x2x103x100)(550-0.42x100) = 731.52 KNm > 464.79 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 Choose 20 diameter bars, Astb = 314 mm2 No of bars =

=4

SHEAR(KN) Qx 0 -154.93 -154.93 -154.93 -6.19 0 -6.19 0

Qy 0 0 0 0 0 0 -18.58 18.58

Provide 4 bars of 20 mm diameter (Ast = 1256 mm2) Shear Reinforcement: Maximum ultimate shear = (154.93) x 2 x 1.5 = 464.79 KN =

v

= 4.22 N/mm2

=

Assuming two bars to b bent up at the supports, Ast at supports = 980 mm2 Pt = c

=

= 0.9

= 0.599 N/mm2 <

Vud

=

v

= (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=

=

(Ast/bd) = pt = Ast

= 1827.23

= 3.93 (1-

)

mm2

For 25 mm diameter bars, no of bars required = 4

Provide 25

st

= 1960 mm2)

For shear reinforcement provid 10 mm diameter bars @ 200 mm c/c SLAB: Mxy = 5.51 KNm Myx = 5.51 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.

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/m2 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

=(2.4x9x12)

=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

=

3) Analysis: a) Section Properties :

=9.94KN/m2

=

= 0.181

=

= 0.1 (X- direction)

=

= 0.133 ( Y-direction)

From design table I = CbwD3 = (0.191x200x5503)

(C=0.191)

= 63.8x108mm4 = I1 =I2 Where I1 and I2 are 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 63.8x108/2x1012)

=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

Assume µ=0.15

; E= 5700

Cy= GC2/a1 = 5700

= 25.49 x 106 KN/m2 G=

= 0.435µ

G= 11.09x106 KN/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

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

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 Span /250

= 26.46 mm

= 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( )2 x0.0088 = 91.049 = Dy ( )2

My = -(Dy )

[

]

= 1.083x105( )2 x0.0088 = 65.25 Mxy = -(

)

= -(

) ( )2

[

]

=(-) (

)(

)x0.0088x

=(-) 4.53 Myx

= -6.05

Qx

=

(Dx

=(-)

+

)

(

)(Dx

+ Dy

)

=(-)0.0088((0.85x105x ) +8.35x103x = -33.52( Qy

)}w1{(

= (-){(0.85x105x ( )3 + 6.265x103( = - 15.57 (

Points

x,y

A B C D E F G H

4.5,6 3,6 1.5,6 0,6 0,10 0,12 4.5,12 4.5,10

)

)

= (-){(Dx( )3 +Cx(

Qy

)) (

) )}x 0.008(

)

)

0 0 0

1 0 0.866 0.5 0.5 0.866 0 1 0 1 0 1 1 0 1 0 Table-4.1

1 1 1 1 0.5 0 0 0.5

0 0 0 0 -0.866 -1 -1 -0.866

POINTS

MOMENTS(KNm) My Mxy 62.5 0 54.125 0 31.25 0 0 0 0 3.92 0 4.53 0 0 31.25 0 Table-4.2

Mx 91.049 78.85 45.524 0 0 0 0 45.524

A B C D E F G H

Myx 00 0 0 0 5.2392 6.05 0 0

5) Design of Reinforcement: Max moment = 91.049 KNm Moment resisted by central rib in x-direction over 1.5m width = 1.5x91.049 = 136.5735 KNm Ultimate moment

= 204.85 KNm

Moment Capacity of Flange section Muf= 0.36fckbfDf(d-0.42Df) = ( 0.36x20x1.5x103x100)(500-0.42x100) = 494.64 KNm > 204.86 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 Choose 20 diameter bars, Astb = 314 mm2 No of bars =

=4

SHEAR(KN) Qx 0 16.76 29.082 33.52 16.76 0 0 0

Qy 0 0 0 0 0 0 15.57 13.48

Provide 4 bars of 20 mm diameter (Ast = 1256 mm2) Shear Reinforcement: Maximum ultimate shear = (33.52) x 2 x 1.5 = 100.56 KN =

v

= 1.0056 N/mm2

=

Assuming two bars to b bent up at the supports, Ast at supports = 628 mm2 Pt = c

=

= 0.628

= 0.528 N/mm2 <

Vud

=

v

= (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 = 62.5 KNm Mu = 2x62.5x1.5 = 187.5 KNm R=

=

(Ast/bd) = pt = Ast

= 0.375 (1-

= 1062.1125

)

mm2

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

Provide 20 mm diameter bars of 4

st

= 1256 mm2)

For shear reinforcement provid 8 mm diameter bars @ 400 mm c/c SLAB: Mxy = 4.53 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.

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/m2 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 = 500 mm

Width of rib

=200mm

Adopt overall depth of ribs

=600mm

2)Loads: 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

Load per meter

=

=9.44KN/m2

3) Analysis: a) Section Properties : =

= 0.167

=

= 0.1 (X- direction)

=

= 0.1 ( Y-direction)

From design table = CbwD3

I

= (0.191x200x6003)

(C=0.191)

= 82.51x108mm4 = I1 =I2 Where I1 and I2 are 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 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

=25.49 x 106 KN/m2 G=

= 0.435µ

G= 11.09x106 KN/m2 C1=C2= (1-0.63(x/y))(x3y/3) = (1 - 0.63(0.2/.6))( C1=C2= 1.264 X10-3

)

= 5700

Cx= GC1/b1 =

= 7.002x103

Cy= GC2/a1 =

= 7.002x103

2H = Cx +Cy = 14.001 x 103m unit 3) Central Deflection: (a= 10m , b=12m) =

= 10.5 ----------------------------------------------- (A)

=

= 5.06 ----------------------------------------------- (B)

=

= .9722 ----------------------------------------------(C)

A+B+C

= 16.53

Central Deflection

(W1)

=

=

= 9.3mm

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 Span /250

= 10000/250

= 27.9 mm

= 40 >26.46

(IS 456-2000)

Maximum deflection is within the permissible limits. 4) Design Moments and Shear: Mx = -(Dx )

=

( )2

[

]

= 0.85x105( )2 x0.0093 = 116.24 My = -(Dy )

= Dy ( )2

[

=1.083x105( )2 x0.0093

]

= 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(

=(-) 15.57 (

Points

x,y

A B C D E F G H

4.5,6 3,6 1.5,6 0,6 0,10 0,12 4.5,12 4.5,10

0 0 0

)

) )}w1{(

= (-) {(0.85x105x ( )3 + 6.265x103( Qy

)) (

) )}x 0.008(

)

)

1 0.866 0.5 0 0 0 1 1

0 0.5 0.866 1 1 1 0 0 Table-4.1

1 1 1 1 0.5 0 0 0.5

0 0 0 0 -0.866 -1 -1 -0.866

POINTS

MOMENTS(KNm) My Mxy 62.5 0 54.125 0 31.25 0 0 0 0 3.92 0 4.53 0 0 31.25 0 Table-4.2

Mx 91.049 78.85 45.524 0 0 0 0 45.524

A B C D E F G H

Myx 00 0 0 0 5.2392 6.05 0 0

5) Design of Reinforcement: Max moment = 116.24 KNm Moment resisted by central rib in x-direction over 2m width = 2x116.24 = 174.24 KNm Ultimate moment = 261.54 KNm Moment Capacity of Flange section Muf= 0.36fckbfDf(d-0.42Df) = ( 0.36x20x2x103x100)(550-0.42x100) = 548.64 KNm > 261.54 KNm Since Mu < Muf Neutral axis falls within the flange Mu = 0.87fyAstd (1-

)

100 (Ast/bd) = pt =

(1-

=

(1-

)

)

= 1.39x 10-3 1396..21 mm2 Choose 25 diameter bars, Astb = 785mm2 No of bars =

=4

SHEAR(KN) Qx 0 16.76 29.082 33.52 16.76 0 0 0

Qy 0 0 0 0 0 0 15.57 13.48

Provide 4 bars of 25 mm diameter (Ast = 3140 mm2) Shear Reinforcement: Maximum ultimate shear = (36.52) x 2 x 2 = 138.08 KN =

v

= 1.218 N/mm2

=

Assuming two bars to b bent up at the supports, Ast at supports = 628 mm2 Pt = c

=

= 0.628

= 0.528 N/mm2 <

Vud

v

=

= (1.218-0.5218) x 200 x 550 = 48.432x103 = 0.2684

Provide 8 mm dia bars 2 legged ,

= 2x50.24 mm2

= 0.2648 = 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=

=

(Ast/bd) = pt = Ast

= 1440.71

= 0.462 (1-

)

mm2

For 22 mm diameter bars, no of bars required = 4

Provide 20 mm

st

= 1519mm2)

For shear reinforcement provid 8 mm diameter bars @ 400 mm c/c SLAB: Mxy = 4.73 KNm Myx = 6.88KNm 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.

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: 1)Dimension of the Slab: 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: a) Section Properties : =

= 0.166

=

= 10(X- direction)

=

= 0.133 ( Y-direction)

From design table I

= CbwD3 = (0.191x200x6003) = 82.6*10^-4M^4 = I1 =I2

(C=0.191)

Where I1 and I2 are 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) =(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 106 KN/m2 = 0.435µ

G =

G= 11.09x106 KN/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 3) Central Deflection: (a= 12m , b=16m) =

= 5.09 ----------------------------------------------- (A)

=

= 1.61 ----------------------------------------------- (B)

= A+B+C

= .38 ----------------------------------------------(C) = 7.08

Central Deflection

(W1)

=

=

= 15 mm

Long term deflection is equal to 2 to 3 times elastic deflection depending on the date of removal of supports. Long term deflection = 2 x 15 Span /250

= 30 mm

= 12000/(2.5*15)

= 320 324KNm Since Mu < Muf Neutral axis falls within the flange Mu = 0.87fyAstd (1-

)

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) Shear Reinforcement: Maximum ultimate shear = (29.52) x 2 x 2 = 118.08 KN =

v

= .984 N/mm2

=

Assuming two bars to b bent up at the supports, Ast at supports = 628 mm2 Pt = c

=

= 0.523

= 0.523 N/mm2 <

Vud

v

=

= (.984-0.523) x 200 x 600 = 55.32x103 = 0.255

Provide 8 mm dia bars 2 legged ,

= 2x50.24 mm2

= 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=

=

(Ast/bd) = pt = Ast

= 2.54 (1-

= 1030.1125

)

mm2

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 = 3.74 KNm Myx = 5.3 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.