GRID FLOORS INTRODUCTION The floor which is resting resting on the beams running running in two directions directions is known as grid floor. In these these types types of of floor, a mesh mesh or grid of beams beams running running in both both the directions directions is the main main structure, structure, and the floor is of nominal nominal thickness. thickness. It is used used to cover a large area without without obstructio obstruction n of internal internal columns. columns. They are generally 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 Typical Plan of Grid System System and Notations 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 urpo urpose se,, the the widt width h of of the the hall hall is kept kept cons cons ant as 10.00 m and and leng length th is incr increa ease sed d by a n int inter erval val of 1.00 1.00 m, m, so so tha thatt L/B L/B rat ratio io var varies from 1 to 1.5 at the interval interval of 0.1. 0.1. For al al l hall sizes the thickness of grid slab is as umed as 100 mm. The size of the beams (0.2 (0.23 3 m x 0.6 0.60 0 m) m) is is kep keptt sam samee dur durin ing g en en ire study.
Analysis Of Grid Floors (a) Approximate Methods According to the Indian St ndard ndard Code Code IS: 456-19 456-1978, 78, the the ribbed ribbed sla sla system can be analys analysed ed as solid solid slab slab,, if th th foll followi owing ng requ requir irem emen ents ts rega regard rdin ing g spac spacii g of beams, thickness of slab and edge eams are satisfied. 1. l. The spacing Of the ribs should not be greater than 1.5 m and it hould not be greater than 12 times the flange thickness. 2. In situ ribs should not be less than 65 mm wide. 3. The The ribs ribs should should be forme formed along each edge paral rallel to the span, having a width equal to that that of the bearin bearin . The moments and shears er unit unit width width Of grid are determ determine ined d fro Table 22 Of IS : 456 code and the reinforcements are designed in the ribs. The slab reinforcement ge eral erally ly cons consis ists ts of a mesh mesh or fabr fabric ic.. A sec 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 re a 1 and b1 in the x and y directions resp ctively. q = Total Total load load per unit unit are and and q1 and and q2 are the load loadss share shared d in t e x and y directions a= Shorter dimension of gr id b= Longer dimension of gri d The deflections of the ribs AB and CD at the junction O must be the same and by equating the deflections, we have have deflect deflection ion = 5
5 =
384
384
= =
solving the above equations we have =
=
and
+
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 following sections, a brief discussion 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 Sti fness Matrix Method This This is an exact exact method method an needs a computer for its application. It a ssumes the frame coordinates system X and and the member coordinate system X' — shown in — Y' as shown Fig. Fig. 6.1(b) 6.1(b).. The activ activee joint joint s are identified first and then the force dis placement equati equations ons are are writte written n for th vari various ous acti active ve membe memberr ends. ends. The The memb member stiffnesses are then then transfe transferre rred d to the the fram fram coor oordina dinate te syst system em.. From From the the stif stiffn fnes esss matrix atrix and joint joint displacement equations, the force force at eac each h membe memberr join jointt can can be be obtai obtain n ed by use of computer programs. Ready ade ade pro progr gram amss for for the the va variou riouss type typess of of g ids are also available for quick analysis in design offices. If a computer and the r levant software programs are available, the analys lysis can be be ca carri rried ou out qu quick ickly. ly. Ho How ver, the results of the computer analysis sh ould ould be checke checked d with with valu values es obta obtaine ined d fro fro approximate methods.
Analysis Of Grid Floors By Equating Joint Deflections In this this meth method od,, the the defl deflec ecti tion of the joints and the loads on each Of he beams are determined. As the number of resulting resulting equations equations is large, large, a compute computer will be required for for its its eas easy y solu soluti tion on.. (Ana (Analy lysis of Grid Floors ) by the National Build ings Organisa isation tion,, Ne New De Delhi lhi gi gives tables and charts which may be conv niently used for the analysis of rectangular nd 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 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 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 10 )/20 )/20 = 450 mm
Width of rib
=200mm
Adopt overall depth of ribs
=550mm
3
2)Loads: 2
Weight of the slab
=(0.1x24)
=2.4KN/m
Total load of slab
=(2.4x9x9)
=194.4KN
Weight of ribs
=(.2x.45x24) =2.16KN/m
Tot Total wei weigh ghtt of beam beam(Y (Y-d -diirect rectio ion) n)
=(7x =(7x2. 2.16 16x9 x9))
=136 =136.0 .08 8KN
Tota Totall wei weigh ghtt of of bea beam( m(XX-di dire rect ctio ion) n)
=(7x =(7x2. 2.16 16x9 x9))
=136 =136.0 .08K 8KN N
Total weight of Floor Finish
=(0.6x9x9)
=48.6KN
Total live load
=(4x9x9)
=324KN
Total (D ( Dead load+Live Load)
=843.72 KN
Load per meter
=
2
=10.416KN/m
3) Analysis: a) Section Properties : =
= 0.181
=
= 0.133 (X- direction)
=
= 0. 0.133 ( Y-direction)
From design table 3
I
= CbwD
3
= (0.191x200x550 ) 8
(C=0.191)
4
= 63.8x 63.8x10 10 mm = I1 =I2 Where I1 and I2 are the second moment of inertia of T-sections about their centrodial axis axis in the x and y directions respectively. D1
= E I1
and
D2 = E I 2 8
12
Dx= (EI1 /b1) =(E x 63.8x10 /1.5x10 )
=0.00425E
Dy= (EI1 /a1) =(E x 63.8x10 8 /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
6
=25.49 x 10 KN/m G=
Cy= GC2 /a1
2
= 0.435µ 6
G = 11. 11.09 09x1 x10 0 KN/m
2
C1= C2 = (1-0.63 (1-0.63(x/y (x/y))(x ))(x3y/3) = (1 - 0.63(0.2/055))( C1=C2= 1.13 X10 -3
)
= 5700
Cx= GC1/b1 =
= 8.35x103
Cy= GC2/a1 =
= 8.35x10
3
3
2H = Cx Cx +Cy = 16.7 x 10 m unit 3) Central Deflection: (a= 9m , b=9m) =
= 12.34 ----------------------------------------------- (A)
=
= 12 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 Long term term defl deflec ecti tion on = 3 x 8.82 .82
= 26.4 26.46 6 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
[
5
]
2
=0.85x10 ( ) x0.0088 = 117.46
My = -(Dy )
2
= Dy ( )
[
=1.083x105( )2 x0.0088
]
= 115.736
Mxy = (-)(
2
) ( )
= ( -) (
[
)(
]
)x0.0088x
= (-) 6.53 Myx
= ( - ) 6 .0 5
Qx
=
= ( -)
(Dx
+
)
(
)(Dx
+ Dy
5
)
= (-) 0.008 0.0088(( 8((0. 0.85 85x10 x10 x ) +8.35x103x = -37.02( Qy
3
= (( -) {( { (Dx( ) +Cx(
)}w1{( 3
= - 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, 1 2 4.5,10
)
) ) 3
=(-) =(-) {(0. {(0.8 85x10 5x105x 5x ( ) + 6.265x10 ( 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 Mx y 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
My x 00 0 0 0 6. 0 5 6. 0 5 0 0
5) Design Design of Reinforcement: Reinforcement: Max Max mome moment nt = 117. 117.46 46 KNm 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 = 0.36f ck ck bf Df (d-0.42Df ) 3
= ( 0.36x20x1 0.36x20x1.5x1 .5x10 0 x100)(500-0.42x100) = 494.64 KNm > 264.86 KNm Since Since Mu < Muf Neutral Neutral axis fall fallss within within the the flange flange Mu = 0.87f yAstd (1-
)
100 (Ast/bd) = pt =
(1-
=
( 1-
= 1.1629x 10
)
) -3
1162.93 mm
2
Choose 20 diameter bars, Astb = 314 mm
2
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 2
Provide 4 bars of 20 mm mm diameter diameter (Ast = 1256 mm ) Shear Reinforcement: Reinforcement: Maximum Maximum ultima ultimate te shear shear = (37.02 (37.02)) x 1.5 x 1.5 = 83.295 83.295 KN =
v
=
= 0.932 N/mm
2
Assuming two bars to b bent up at the supports, A st at supports = 628 mm Pt = c
=
= 0.628 2
= 0.528 N/mm <
Vud
=
v
= (1.0056-0.5218) x 200 x 500 = 48.432x103 = 0.26 0.2684 84
Provi rovide de 8 mm mm di dia bar barss 2 leg legged ,
= 2x 2x50. 50.24 mm2
= 0.2648 = 374.255 3d
=3x500 = 1500m
Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Along YY- Direction: Direction: M =115.736KNm Mu = 2x115.736x1.5 =326.205 KNm R=
=
(Ast /bd) = pt = Ast
= 0.655 (1-
= 1062.1125
mm
) 2
2
For 20 mm diameter bars, no of bars required = 4 st
= 125 1256 6 mm2)
For shear reinforcement provid 8 mm diameter bars @ 400 mm c/c SLAB: Mxy = 6.0 KNm 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 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. 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
= (10x10 )/20 )/20 = 500m 500mm m
Width of rib
=200mm
Adopt overall depth of ribs
=600mm
3
2)Loads: 2
Weight of the slab
= (0.1x24)
=2.4KN/m
Total load of slab
= (2 (2.4x10x14) =3 =336KN
Weight of ribs
= (.2x.6x24)
Total Total weig weight ht of beam( beam(YY-di direc recti tion on))
= (8x2 (8x2.88 .88x1 x14) 4) =322. =322.56 56KN KN
Total Total weig weight ht of beam( beam(XX-di direc recti tion on))
= (6x2 (6x2.88 .88x1 x10) 0) =172. =172.8K 8KN N
Total weight of Floor Finish
= (0.6x10x14) =84KN
Total live load
= (4x10x14)
=2.88KN/m
=560KN
Total (Dead load+Live Load)
= 14 1 475.36KN
Load per meter
=1475.36/(10x14) = 10.53KN/m
2
3) Analysis: a) Section Properties : D /D f w =100/600
= 0.166
bw /bf =200 =200/2 /2000 000 = 0.1 (X(X- direc directi tion on)) =200/2000 =200/2000 = 0.1 ( Y-direction) Y-direction) From design table I
3
= CbwD
3
= (0.191x200x600 )
(C=0.191)
= 82.5x 82.5x10 108mm4 = I1 =I2 Where I1 and I2 are the second moment of inertia of T-sections about their centrodial axis axis in the x and y directions respectively. D1
= E I1
and
D2 = E I 2 8
12
=0.00319E
8
12
=0.00319E
Dx= (EI1 /b1) =(E x 82.5x10 /2x10 ) Dy= (EI1 /a1) =(E x 82.5x10 /2x10 )
b) The Torsional Rigidity Rigidity in X and Y directions are given by Cx=GC1 /b1
and
Cy= GC2 /a1
Assume µ=0.15
; E= 5700 6
= 5700
2
=25.49 x 10 KN/m G= (E/2(1+ µ)) = 0. 0.435µ 435µ 6
G= 11.09x10 KN/m
2
C1=C2= (1-0.63(x/y))(x3y/3) 3
= (1 - 0.63(0. 0.63(0.2/0. 2/0.55)) 55))((0. ((0.2 2 x0.55)/3) C1=C2= 1.13 X10
-3
6
-3
3
Cx= GC1/b1 GC1/b1 = (((11.09x10 (((11.09x10 )(1.13x10 ))/2)= 6.26x10
Cy= GC2/a1 GC2/a1 = (((11.09x10 (((11.09x106)(1.13x10-3))/2)= 6.26x103 3
2H = Cx +Cy +Cy = 12.52 12.52 x 10 m unit
3) Central Deflection: (a= 10m , b=14m) (Dx /a4)
= (( ((105x0.81)/104) = 8.1 ------------------------------------------------------------------------------------------ (A)
4
5
(Dy /b )
4
= ((10 x0.81)/14 ) =2.10 =2.10 ------------------------------------------------------------------------------------ (B) 2 2
3
2
2
((2H)/(a b )) =((1 =((12. 2.52 52x1 x10 0 )/(10 14 ))= 0.638 -------------------------------------------------------------------------------------(C) ----(C) A+B+C
= 10.83
Central Deflection
(W1) = ((16q)/(1 ((16q)/(10.83 0.83
6
)) = ((16 (16x10 x10.53 .53)/( )/(10.8 0.83
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 Long term term defle deflect ctio ion n = 2 x 16.2 16.2 Span /250
= 32.4 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 )
[
5
]
2
= 0.81x1 0.81x10 0 ( ) x0.0194 = 154.93
My = ((-) (D (Dy )
= ((-) Dy Dy (
2
)
[
]
= (-) (-) 0.81 0.81x1 x10 05( )2 x0.0194 = (-) 79. 79.04
Mxy = (-)(
)
= -(
= (- ) (
) ( )2
[
)(
]
2
) x0.0194x
= -5.5 -5.51 1 Myx
= -5.5 -5.51 1
Qx
=
= (- )
(D (Dx
+
)
(
)(Dx
5
Qy
+ Dy
= (-) 0.01 0.0194 94((0 ((0.8 .81x1 1x10 0x
) +48.02x
= (-)154.93(
) 3
= (-) {(Dx( ) +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. {(0.81 81x1 x10 05x ( )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. 0 4 0 0. 0 4 0
0 0 0 0 1 -1 1 -1
POINTS
MOMENTS(KNm) My Mx y 79.04 0 1. 6 5 0 0. 7 9 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
My x 00 0 0 0 -5.51 5. 5 1 0 0
5) Design of Reinforcement: Max Max mome moment nt = 154. 154.93 93 KNm KNm Moment resisted by central rib in x-direction over 2m width = 2x154.93 2x154.93 = 309.86 KNm Ultimate mo moment
= 46 464.79 KNm
Moment Capacity of Flange section Muf = 0.36f ck ck bf Df (d-0.42Df ) 3
= (0.36x20 (0.36x20x2x1 x2x10 0 x100)(550-0.42x100) = 731.52 KNm > 464.79 KNm Since Mu < Muf Neutral axis falls within within the flange flange Mu = 0.87f yAstd (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 (A st = 1256 mm2) Shear Reinforcement: Maximum ultimat ultimatee shear = (154.93) x 2 x 1.5 = 464.79 KN =
v
= 4.22 N/mm2
=
2
Assuming Assuming two bars to b bent up at the supports, supports, Ast at suppo supports rts = 980 980 mm Pt = c
=
= 0 .9
= 0.599 0.599 N/mm N/mm2 <
Vud
v
=
= (4.22-0.599) x 200 x 550 = 398.31Sx10
3
= 1.96 Provi rovide de 10 mm dia bars 2 legged ged ,
= 2x157 157 mm2
= 1.96 = 160 3d
=3x550 = 1650m
Provide 2 legged legged 10 mm diameters diameters bars @ 200mm c/c. Mu Along Along YY- Direction: Direction: M = 79.4 79.4 KNm KNm Mu = 2x79.4x1.5 2x79.4x1.5 = 238.2 KNm R=
=
(Ast /bd) = pt = Ast
1827.23 23 = 1827.
= 3. 9 3 (1-
) 2
mm
For 25 mm diameter diameter bars, no of bars bars required = 4
Provide 25
st
= 196 1960 0 mm2)
For shear reinforcement provid 10 mm diameter bars @ 200 mm c/c SLAB: Mxy = 5.5 5.51 1 KNm KNm Myx = 5.5 5.51 1 KNm 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 concrete grid floor of size 9m x 12m is 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 Solution: 1)Dimension of the Slab: Adopt thickness of slab
=100mm
Depth of Ribs: Span/Depth
= 20 (Simply supported case)
Depth
= (9 x 10 )/20 = 450 mm
Width of rib
=200mm
Adopt overall depth of ribs
=550mm
3
2)Loads: 2
Weight of the slab
=(0.1x24)
=2.4KN/m
Total load of slab
=(2.4x9x12)
=259.24KN
Weight of ribs
=(.2x.45x24) =2.16KN/m
Total Total weig weight ht of beam( beam(YY-di direc recti tion on))
=(7x2 =(7x2.1 .16x1 6x12) 2) =181. =181.44 44KN KN
Tota Totall wei weigh ghtt of of bea beam( m(XX-di dire rect ctio ion) n)
=(7x =(7x2. 2.16 16x9 x9))
=136 =136.0 .08K 8KN N
Total weight of Floor Finish
=(0.6x9x12)
=64.8KN
Total live load
=(4x9x12)
=432KN
Total (Dead load+Live Load)
=1073.2 KN KN
Load per meter
=
3) Analysis: a) Section Properties :
2
=9.94KN/m
2
=
= 0.181
=
= 0.1 (X- direction)
=
= 0. 0.133 ( Y-direction)
From design table I = CbwD
3
= (0.191x200x5503) 8
(C=0.191)
4
= 63.8 63.8x10 x10 mm = I1 =I2 Where I1 and I2 are the second moment of inertia inertia of T-sections about their centrodial axis in the x and y directions respectively. D1
= E I1
and
D2 = E I 2 8
12
Dx= (EI1 /b1) =(E x 63.8x10 /2x10 )
=0.00319E
Dy= (EI1 /a1) =(E x 63.8x10 8 /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 6
= 25.49 25.49 x 10 KN/m
Cy= GC2 /a1 = 5700
2
= 0.435µ
G= 6
G= 11.09x10 KN/m
2
3
C1=C2= (1-0.63(x/y))(x y/3) = (1 - 0. 0 .63(0.2/055))( C1=C2= 1.13 X10
)
-3
Cx= GC1/b1 =
= 6.26x10
Cy= GC2/a1 =
= 8.35x10
3
3
2H = Cx +Cy = 14.61 x 10 3m 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
= 90 9000/250
= 26.46 mm
= 36 >26.46
(IS 456-2000)
Maximum deflection is within the permissible limits. 4) Design Moments and Shear: Mx = -(Dx )
=
2
( )
[
5
]
2
= 0.85x1 0.85x10 0 ( ) x0.0088 = 91.049 2
My = -(Dy )
= Dy ( )
[
]
= 1.08 1.083x1 3x10 05( )2 x0.0088 = 65.25
Mxy = -(
)
= -(
2
) ( )
[
]
=(-) (
)(
)x0.0088x
=(-) =(-) 4.53 4.53 Myx
= -6.0 -6.05 5
Qx
=
(D (Dx
=(-)
+
)
(
) ( Dx
+ Dy
)
5
=(-)0.0088((0.85x10 x ) +8.35x103x = -33.52( Qy
)}w1{( 3
) 3
= (-) (-){(0. (0.85x105x 105x ( ) + 6.265x10 ( = - 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, 1 2 4 . 5, 1 0
)
)
= (-){(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 Mx y 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
My x 00 0 0 0 5.2392 6. 0 5 0 0
5) Design of Reinforcement: Max mome moment nt = 91. 91.04 049 9 KNm KNm Moment resisted resisted by central rib in x-direction x-direction over 1.5m width width = 1.5x91.049 = 136.5735 KNm Ult Ultimat imatee mom momen entt
= 204 204.8 .85 5 KNm KNm
Moment Capacity of Flange section Muf = 0.36f ck ck bf Df (d-0.42Df ) 3
= ( 0.36x20x1 0.36x20x1.5x .5x10 10 x100)(500-0.42x100) = 494.64 KNm > 204.86 KNm Since Mu < Muf Neutral axis falls within within the flange Mu = 0.87f yAstd (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 (A st = 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, A st at supports = 628 mm Pt = c
=
= 0.628
= 0.528 N/mm 2 <
Vud
=
v
= (1.0056-0.5218) x 200 x 500 = 48.432x10
3
= 0.2684 Prov Provid idee 8 mm dia dia bars bars 2 legg legged ed ,
= 2x50 2x50.2 .24 4 mm2 mm2
= 0.2648 = 374.255 3d
=3x500 = 1500m
Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Along YY- Direction: Direction: M = 62.5 KNm Mu = 2x62.5x1.5 = 187.5 KNm R=
=
(Ast /bd) = pt = Ast
= 0.375 (1-
= 1062.1125
mm
) 2
For 20 mm diameter bars, no of bars required = 4
2
Provide 20 mm diameter bars of 4
st
= 125 1256 6 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 concrete and 2
Fe415 grade steel. Live load may be assumed as 4KN/m Solution: 1)Dimension of the Slab: Adopt thickness of slab
=100mm
Depth of Ribs: Span/Depth
= 20 (Simply supported case)
Depth
= (9 x 10 )/20 = 500 mm
Width of rib
=200mm
Adopt overall depth of ribs
=600mm
3
2)Loads: 2
Weight of the slab
=(0.1x24)
=2.4KN/m
Total load of slab
=(2.4x9x12)
=288KN
Weight of ribs
=(.2x.45x24) =2.16KN/m
Total Total weig weight ht of beam( beam(YY-di direc recti tion on))
=(7x2 =(7x2.1 .16x1 6x12) 2) =120K =120KN N
Tota Totall weig weight ht of beam beam(X (X-d -dir irec ecti tion on))
=(7x =(7x2. 2.16 16x9 x9))
=172 =172KN KN
Total weight of Floor Finish
=(0.6x9x12)
=72KN
Total live load
=(4x9x12)
=480KN
Total (Dead load+Live Load)
=1132.8 KN KN
Load per meter
=
2
=9.44KN/m
3) Analysis: a) Section Properties : =
= 0.167
=
= 0.1 (X- direction)
=
= 0. 0 .1 ( Y-direction)
From design table 3
I
= CbwD
3
= (0.191x200x600 ) 8
(C=0.191)
4
= 82.51 82.51x1 x10 0 mm = I1 =I2 Where I1 and I2 are the second moment of inertia of T-sections about their centrodial axis axis in the x and y directions respectively. D1
= E I1
and
D2 = E I 2 8
12
Dx= (EI1 /b1) =(E x 82.51x10 /2x10 )
=0.00412E
Dy= (EI1 /a1) =(E x 82.51x10 8 /2x1012)
=0.00412E
b) The Torsional Rigidity in X and Y directions are given by Cx=GC1 /b1
and
Assume µ=0.15
; E= 5700 6
Cy= GC2 /a1
=25.49 x 10 KN/m
2
= 0.435µ
G= 6
G= 11.09x10 KN/m
2
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 =
= 7.002x10
Cy= GC2/a1 =
= 7.002x10
3
3
3
2H = Cx +Cy = 14.001 x 10 m unit 3) Central Deflection Deflection:: (a= 10m , b=12m) =
= 10.5 ----------------------------------------------- (A)
=
= 5.06 ----------------------------------------------- (B)
=
= .9722 ----------------------------------------------(C)
A+B+C
= 16.53
Central Deflection
(W1)
=
=
= 9 .3 m 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 Span /250
= 10 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
[
5
]
2
= 0.85x1 0.85x10 0 ( ) x0.0093 = 116.24
My = -(Dy )
2
= Dy ( )
[
5
]
2
=1.083x10 ( ) x0.0093
= 67.25
Mxy = -(
)
= -(
2
) ( )
= (-) (
)(
[
]
)x0.0088x
= -4.7 -4.73 3 Myx
= -6.8 -6.85 5
Qx
=
= (- )
(D ( Dx
+
)
(
)(Dx
+ Dy
)
5
= (-) 0.009 0.00938( 38((0. (0.85 85x1 x10 0 x ) +8.35x103x = (-) 36.122( Qy
3
= (-){(Dx( ) +Cx(
)}w1{( 3
=(-) 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, 1 2 4 . 5, 1 0
0 0 0
)
) ) 3
= (-) {(0.85x105x ( ) + 6.265x10 ( 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 Mx y 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
My x 00 0 0 0 5.2392 6. 0 5 0 0
5) Design of Reinforcement: Max mome moment nt = 116 116.2 .24 4 KNm KNm Moment resisted by central rib in x-direction over 2m width = 2x116.24 = 174.24 KNm Ulti Ultima mate te moment moment = 261.5 261.54 4 KNm KNm Moment Capacity of Flange section Muf = 0.36f ck ck bf Df (d-0.42Df ) 3
= ( 0.36x20 0.36x20x2x x2x10 10 x100)(550-0.42x100) = 548.64 KNm > 261.54 KNm Since Mu < Muf Neutral axis falls within within the flange Mu = 0.87f yAstd (1-
)
100 (Ast/bd) = pt =
(1-
=
( 1-
)
)
= 1.39x 10-3 1396..21 mm2 Choose 25 diameter diameter bars, 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 Provide 4 bars of 25 mm diame diameter ter (A (Ast = 314 3140 0 mm2) Shear Reinforcement: Maximum ultimat ultimatee 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, A st at supports = 628 mm Pt = c
=
= 0.628
= 0.528 N/mm 2 <
Vud
v
=
= (1.218-0.5218) x 200 x 550 = 48.432x10
3
= 0.2684 Prov Provid idee 8 mm dia dia bars bars 2 legg legged ed ,
= 2x50 2x50.2 .24 4 mm2 mm2
= 0.2648 = 262.91 3d
=3x550 = 1650m
Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Along YY- Direction: Direction: M = 67.5 KNm Mu = 2x67.5x2= 270 KNm R=
=
(Ast /bd) = pt = Ast
= 1440.71
= 0.462 (1-
) 2
mm
For 22 mm diameter bars, no of bars required = 4
2
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 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 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: 2
Load Load per per met meter er =6.5 =6.5KN KN/m /m 3) Analysis: a) Section Properties : =
= 0.166
=
= 10(X- direction)
=
= 0. 0.133 ( Y-direction)
From design table I
3
= CbwD
3
= (0.191x200x600 ) = 82.6*10 82.6*10^-4M ^-4M^4 ^4 = I1 =I2
(C=0.191)
Where I1 and I2 are the second moment of inertia of T-sections about their centrodial axis axis in the x and y directions respectively. D1
= E I1
and
D2= EI EI2
Dx= (EI1 /b1) =(25.49*10^6 x 82.8x10 /2x10 )
8
12
=1.055*10^5M
Dy= (EI1 /a1) =(25.49*10^6x 82.8x10 8 /2x1012)
=1.055*10^5M
b) The Torsional Rigidity in X and Y directions are given by Cx=GC1 /b1
and
Assume µ=0.15
; E= 5700
6
=25.49 x 10 KN/m G =
Cy= GC2 /a1 = 5700
2
= 0.435µ 6
G= 11.09x10 KN/m
2
3
C1=C2= (1-0.63(x/y))(x y/3) = (1 - 0.63(0.2/0.6))(
)
-3
C1=C2= 1.264 X10
3
Cx= GC1/b1 =
= 7x10
Cy= GC2/a1 =
= 7x103 3
2H = Cx +Cy = 14 x 10 m 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 <250(allowed)
(IS 456-2000)
Maximum deflection is within the permissible limits. 4) Design Moments and Shear: Mx = -(Dx )
( )2
=
[
5
]
2
=0.85x10 ( ) x0.0088 = 91.049 2
My = -(Dy )
= Dy ( )
[
5
]
2
=1.083x10 ( ) x0.0088 = 65.25
Mxy = -(
)
= -(
2
) ( )
= (- ) (
[
)(
]
)x0.0088x
= (-)4.5 (-)4.53 3 Myx
= -6.0 -6.05 5
Qx
=
(Dx
= (- )
+
(
)
) ( Dx
+ Dy
)
5
= -0.0088 -0.0088((0. ((0.85x1 85x10 0 x ) +8.35x103x = -33.52( = -{(Dx( ) +Cx(
)}w1{(
)
=-{( =-{(0. 0.85 85x1 x105 05x x ( )3 + 6.265x103( Qy
= - 15.57 (
Points
x ,y
A B C D E F G H
68 4 ,8 2 .8 , 0 ,8 0,12 0,16 6,16 0 . 5, 1 0
POINTS
A B C D E F G H
MOMENS (KNm) Mx 108 94 54 0 0 0 0 77
0 0 0
SHEAR (KN) My 61 53 30.5 0 0 0 0 43
)}x 0.008(
)
)
1 0.866 0 .5 0 0 0 1 1 Table-4.1
0 0. 5 0.866 1 1 1 0 0
POINTS
A B C D E F G H Table-4.2
1 1 1 1 0. 5 0 0 0. 5
MOMENTS (KNm) Mx 108 94 54 0 0 0 0 77
5) Design of Reinforcement: Max Max mom momen entt = 108 108 KNm KNm Moment resisted by central rib in x-direction over 2m width = 2x108 = 216 KNm Ultimate moment
)
) 3
Qy
)) (
= 324 KNm
SHEAR (KN) My 61 53 30.5 0 0 0 0 43
0 0 0 0 -0.866 -1 -1 -0.866
POINTS
A B C D E F G H
Moment Capacity of Flange section Muf = 0.36f ck ck bf Df (d-0.42Df ) =( 0.36x20x2x103x100)(500-0.42x100) = 659 KNm > 324KNm Since Mu < Muf Neutral axis falls within within the flange flange Mu = 0.87f yAstd (1-
)
100 (Ast/bd) = pt =
(1-
=
( 1-
= 2078x 10
)
) -3
2078 mm2 2
Choose 25 diameter bars, Astb = 490.8mm No of bars =
= 4. 2 3
Provide 4 bars of 25 mm and 1 bar of 20 mm diameter (A st = 2277 mm2) Shear Reinforcement: Maximum ultimate shear = (29.52) x 2 x 2 = 118.08 KN =
v
=
= .984 N/mm
2
Assuming Assuming two bars to b bent up at the supports, supports, Ast at supports = 628 mm Pt = c
=
= 0.523 2
= 0.523 N/mm <
Vud
v
=
= (.984-0.523) x 200 x 600 = 55.32x103 = 0.255
Prov Provid idee 8 mm dia dia bars bars 2 legg legged ed ,
= 2x50 2x50.2 .24 4 mm2 mm2
2
= 0.255 = 394.039 3d
=3x500 = 1500m
Provide 2 legged 8mm diameters bars @ 400mm c/c. Mu Along Along YY- Direction: Direction: M = 61KNm Mu = 2x61x1.5 = 183.5 KNm R=
=
(Ast /bd) = pt = Ast
= 2. 5 4 (1-
= 1030.1125
)
mm2
For 20 mm diameter bars, no of bars required = 4 st
2
= 125 1256 6 mm )
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.
