1. Using partial safety factor for loads in accordance with clause 36.4 of IS-456 2000 as ϒt=1.5
2. Partial safety factor for material in accordance with clause 36.4.2 is IS-456-2000 is taken as 1.5 for concrete and 1.15 for steel.
3. Using partial safety factors in accordance with clause 36.4 of IS-456-2000 combination of load.
iii) Flooring material(c.m) 1.0KN/m2
Using M25 and Fe 415 grade of concrete and steel for beams, slabs, footings, columns.
1 Slab is assumed to be continuous over interior support and partially fixed on edges, due to monolithic construction and due to construction of walls over it.
2 Beams are assumed to be continuous over interior support and they frame in to the column at ends.
5.4.1 Assumptions on design:
1. M25 grade is used in designing unless specified.
2. Tor steel Fe 415 is used for the main reinforcement.
3. Tor steel Fe 415 and steel is used for the distribution reinforcement.
4. Mild steel Fe 250 is used for shear reinforcement.
5.5 Data Collection
The building models are 15 storey’s located in zone II. Tables 4.0 and Table 4.2present a summary of the building parameters.
Table 3: General data collection and condition assessment of building
Sl.No
----CHAPTER-6
DESIGNING 6.1 Flat Slab
Figure 12 : flat slab representation
Figure 13 : flat slab sizes Given data:
Interior panel = 12.11 x 10.21 m Live load = 4 KN/ m2
Floor finished load = 1 KN/ mm2 fck = 25 N/mm2
fy = 415 N/mm2
Column size = 750 x 750 mm Thickness of slab:
Thickness of slab = 40, if mild steel = 32, if fy415 or fy500 Thickness of slab, d = span/32
= 12116/32
Clear span = 12.116 - 0.75 = 11.366 m
Mulimit = 0.138fckbd2
Ks = 0.5 + c
S = (/4 x 162 / 14390) x 10211
Width = 10211 mm
Using 16mm bar spacing requirement is S = (/4 x 162 / 6893.37) x 10211 = 297.8 mm
Provide 16mm bars at 300 mm C/C.
For positive moment middle strip:
Mu=790.784x106KNmm d = 380 mm
Mu = 0.87fyAstd [1-Astfy/bdfck]
790.78x106 = 0.87 x 415 x Ast x 380 [1 - Astx415 / 10211x380x25]
= 137199Ast [1 – 20.75Ast / 97x106] Ast = 5913.4 mm2
Using 16mm bar spacing requirement is
S = (/4 x 162 / 5913.4) x 10211 = 347.2 mm Provide 16mm bars at 400 mm C/C.
6.2 Design of Beams
Figure 14 : Beams representation
Figure 15 : Beams sizes
Figure 16 : Beams Area
Figure 17 : B.M Diagram
6.2.1 Members - B31, B32 & B39 (from ETABS)
B39:
Figure 18 : Area B39
Size of beam = 230 x 650 mm.
Effective cover = 25 mm.
Effective depth (d) = 650-25 = 625 mm.
Using M25 grade concrete and Fe 415 grade steel.
Required area of beam is shown in fig. below 3252
3487
╠
Hence provide 3 no.s of main bars and 4 no.s of extra bars at sides.
Required area of steel (whole bottom of beam); Ast = 2468 mm2 (from Etabs)
Provided area of steel; ast
Provide 25 mm dia. Bar; ast = П x 252 / 4
Hence provide 3 no.s of main bars and 2 no.s of extra bars at sides.
Required area of steel (top right of beam); Ast = 3487 mm2
Hence provide 3 no.s of main bars and 4 no.s of extra bars at sides.
Figure 19 : Steel Detail - B39
B31:
Figure 20 : Area B31
Size of beam = 230 x 450 mm.
Effective cover = 25 mm.
Effective depth (d) = 450-25 = 425 mm.
Using M25 grade concrete and Fe 415 grade steel.
Required area of beam is shown in fig. below
248 397
╠
╣
261
Number of bars in a beam = Ast / ast
= 261 / 113.1
= 2.3 ≈ 3 No’s
Hence provide 2 no.s of main bars and 1 no.s of extra bars.
Required area of steel (top right of beam); Ast = 397 mm2 (from Etabs)
Provided area of steel; ast
Provide 12 mm dia. Bar; ast = П x 122 / 4
= 113.1 mm2
Number of bars in a beam = Ast / ast
230
= 397 / 113.1
=3.5 ≈ 4 No.s 450
Hence provide 2 no.s of main bars and 2 no.s of extra bars.
Figure 21 : Steel Detail - B31
B32:
Figure 22 : Area B32
Size of beam = 230 x 450 mm.
Effective cover = 25 mm.
Effective depth (d) = 450-25 = 425 mm.
Using M25 grade concrete and Fe 415 grade steel.
Required area of beam is shown in fig. below
377 248
╠
╣
248
Number of bars in a beam = Ast / ast
= 248 / 113.1
= 2.2 ≈ 3 No.s
Hence provide 2 no.s of main bars and 1 no.s of extra bars.
Required area of steel (top right of beam); Ast = 248 mm2 (from Etabs)
Provided area of steel; ast
Provide 12 mm dia. Bar; ast = П x 122 / 4 230
= 113.1 mm2
Number of bars in a beam = Ast / ast
450
= 248 / 113.1
= 3.2 ≈ 3 No.s
Hence provide 2 no.s of main bars and 1 no.s of extra bars.
Figure 23 : Steel detail - B32
6.3 Column
Figure 24 : Center line Diagram
Figure 25 : Columns representation
Figure 26 : Columns B.M
Column name: C25
Figure 27 : Area C25
Given data:
Column size = 381 x 229 mm M25 grade of concrete
Fe 415 grade of steel
Reinforcement is distributed equally on all four sides Factored axial load, Pu = 17.97 kN. (From Etabs) Factored moment about X-axis, Mux = 10.69 kN-m. (From ETABS)
Factored moment about Y-axis, Muy = 4.99 kN-m. (from ETABS)
Area of steel ; Ast = 697 mm2 Effective length:
Assuming effectively held in position and restrained against rotation at one end , and at the other restrained against rotation but not held in position.
M25 grade of concrete Fe 415 grade of steel
Reinforcement is distributed equally on all four sides Factored axial load, Pu = 2863.98 kN. (from Etabs) Factored moment about X-axis, Mux = 80.66 kN-m. (from rotation at one end , and at the other restrained against rotation but not held in position.
= 5161 / 490.8 Factored axial load, Pu = 7700.45 kN. (From Etabs) Factored moment about X-axis, Mux = 75.6 kN-m. (From rotation at one end , and at the other restrained against rotation but not held in position.
Unsupported length of column (L) = 3000 mm D = 762 mm
3000 / 762 = 3.9 Slenderness ratio, Le/b = < 12 Therefore, it is designed as short column
Provide 25 mm dia. Bar; ast = П x 252 / 4 = 490.8 mm2
Number of bars in a column = Ast / ast
762
= 5161 / 490.8
=10.5≈12No.s 762
Hence provide 6 no.s of bars at one side and 6 no.s of bars at other side.
Figure 28 : Steel Detail - C5, C12 & C25
6.5 Footing
B = 3.24 x 1
= 3.24 m
A = 3.5 x 3.5 = 12.25 mm2
Net upward soil pressure Foundation = 1.5 x 2863.98 / (3.5 x 3.5) = 350.7 KN/m2
Therefore footing design for a maximum pressure of 350kN/m2 4) Determination of minimum depth required from B.M:
Therefor provide an effective depth of footing shear consideration up to two times as 500 mm
Assume effective cover = 50 mm
Overall depth = 500 + 50 = 550 mm
5) Check for one way shear:
Critical section for one way shear is‘d’ from face of the column.
Shear Force Vu = Qu (l - d) x B
7) Area Of Reinforcement Along X and Y Direction:
d = 550 -16 / 2 - 50 = 492mm
Mu = .87fy Ast [d- (fy Ast / bfck )]
1035 x 103 = 0.87 x 415 x Ast [500 – (415 Ast / 3500 x 25)]
Ast = 5453.23 mm2
Provide 16mm ϕ bars, area of single bar = 201.06 mm2 Spacing of bars = (201.6 / 5453.23) x 3500= 129.4 mm Provide 16mm ϕ bars at a spacing of 130 mm c/c.
6.6 Stair Case
Design of Flight Slab:
No of flights for each floor = 2 Height of the floor = 3.05 m
Height of each flight = 3.05 / 2 = 1.53m
Assume Raise as 150mm and Tread as 300mm No of Raisers = 1.53 / 0.15 = 10.2 Say 10 Say Raise = 155mm
No of Treads = 10-1 = 9 Treads Width of Stair = 1.68m
1.68m 1.68m
2.64m 2.81m
Effective Span = 5.22m
Span / Overall depth = 20 for deflection criteria
Let modification factor = 5220 / (25) = 208mm Say 200mm Effective depth = 200-25-5 = 170mm Dead load of steps / horizontal meter run = 0.56/0.3 = 1.87 KN/m Live load 3 KN / m2 = 3.00 KN/m
Total working load = 8.85 KN/m
0.50 x 25 4.6 x 15.9 x 106
Ast = 1-x 1000 1-x 170
415 25 x 1000 x 1702
= 266.09 mm2 > 204 mm2 (Min.Ast 0.12% of bD as per Clause 26.5.2.1
of IS 456-2000)
Provide 12 mm dia. Tor steel ast = П x 122 / 4 = 113 mm2 ast
Spacing of Steel = X 1000 Ast
113
Spacing of Steel = x 1000 = 424.6 mm 266.09
But maximum spacing of Tension Reinforcement as per Clause 26.3.3 b 1 from
IS 456 – 2000 is 3d or 300 mm whichever is less, 1) 3d = 3 x 170 = 510 mm or 2) 300 mm
Hence provide 12 mm dia. @ 300 mm C/C
Revised Astxm = (1000 x 170) / 300 = 566.6 mm2
Distribution Steel:
Area of steel = 0.12% of gross sectional area = (0.12 x 1000 x 170) / 100 = 204 mm
Provide 8 mm dia. Tor steel ast = П x 82 / 4 = 50.26 mm2 ast
Spacing of Steel = X 1000 Ast
50.26
Spacing of Steel = x 1000 = 246.3 mm 204
But maximum spacing of Tension Reinforcement as per Clause
Tension bars crossing at bends should be extended by 773.67mm beyond crossing point.
8) Check for Serviceability:
Basic l/d Ratio = 20
Pt = 0.32% Service stress = (0.58 x 415 x 270)/ 560 = 116N/mm2
Modification factor = 1.20
Modified value of l/d ratio = 20 x 1.20 = 24 Actual l/d ratio = 3100/170 = 18 < 24
Actual l/d ratio is < modified value of l/d ratio Hence the thickness of the slab is safe.
CHAPTER-7
CONCLUSIONS
Flat-slab building structures possesses major advantages over traditional slab-beam-column structures because of the free design of space, shorter construction time, architectural –functional and economical aspects. Because of the absence of deep beams and shear walls, flat-slab structural system is significantly more flexible for lateral loads then traditional RC frame system and that make the system more vulnerable under seismic events.
The purely flat-slab RC structural system is considerably more flexible for horizontal loads than the traditional RC frame structures which contributes to the increase of its vulnerability to seismic effects.
The critical moment in design of these systems is the slab-column
connection, i.e., the penetration force in the slab at the connection, which should retain its bearing capacity even at maximal displacements. The ductility of these structural systems is generally limited by the deformability capacity of the column-slab connection. To increase the bearing capacity of the flat-slab structure under horizontal loads, particularly when speaking about seismically prone areas and limitation of deformations, modifications of the system by adding structural elements are necessary.
REFERENCES
Design of Reinforced Concrete Structures by A. K. Jain,
Surya Prakash S. Krishna Murthy,
Design of R.C.C structural elements by S.S. Bhavikatti,
Design of R.C.C slabs by K.C.Jain,
Design of R.C.C structures by prof.N.Krishna Raju,
Design of R.C.C structures by prof.S.Ramamrutham,
The code books referred for this project are:
1.
2. SP 16 (design aids for IS 456),
3. IS 1893 (part-1) 2002 criteria for earthquake resistant design of structures Part-1 general provision & buildings,
4. IS 456:2000 for RCC design,
5. IS 875 – part I for weight and density of materials ( RCC = 25 kN/m3, PCC = 24 kN/m3, Brick = 18-20 kN/m3 ect.),
6. IS 875 – part II for live load,
7. IS 875 – part III for wind load and 8. IS 875 – part V for load combination.