Two Way Simply Support Slab Calculation /Design
Important Point
Two Way Simply Support slab Below Point Calculation Required
1. Effective Depth (d)
2. Effective Span
3. Load Calculations
4. Mid Span Moment
5. Effective Depth of Flexure
6. Reinforcement in Mild Strip
7. Check for Cracking
8. Check for Deflection
9. Check for Development Length
Effective Depth (d)
For deflection control
L/d = 35 X M.F X 0.8
M.F. Modifiction factor from— IS: 456, p.38.Fig-4
Assume % steel 0.3 to 0.6%
Fs = 0.58 Fy X (Ast requierd / Ast Provied)
Initially assume that Ast reqierd = Ast Provided
Fy = 250 N/ Sq.mm —– Fs = 0.58 X 250 = 145 N/ sq.mm.
Fy = 415 N/ Sq.mm —– Fs = 0.58 X 415 = 240 N/ sq.mm.
Fy = 500 N/ Sq.mm —– Fs = 0.58 X 500 = 290 N/ sq.mm.
Effective Span
• ClearSpan + d
• c/c of Supports
Whichever is smaller ——– as per IS 456-2000 P. 34, CI 22.2.a
Load Calculations
Total Load = D.L. + F.L. + L.L.
Dead load of slab = (d x 25)
Floor Finishing load = (as floor finishing near 1 kn/sq.mm)
Live load = ( as per calculation)
Factor Load = 1.5 x Total load
Mid Span Moment
Corners not held down conditions is given as per IS: 456-2000 P-90 CI D-2
Mx = ax . w . lx . lx
My = ay . w . lx. lx
ax and ay coefficientare obtained from IS: 456 table -26, fig 10.3 shoe nine separate possible arrangement for two way restrained slab.
Effective Depth of Flexure
Mu = 0.138 . fck . b.d.d
Heaer find d
Mu = Sp 16 P 10 Table C
Fck = strength of concrete
b = 1 m area required load
Reinforcement in Mild Strip
Along Lx
- Pt= 50 (fck/fy)x( 1-√(1-(4.6xMu/Fck b.d.d)))
- fck = strenth of concrete
- fy = 415 N/Sq.mm
- Mu = Sp 16 P 10 Table C =0.138 . fck . b.d.d= Mx
Along Lx
- Pt= 50 (fck/fy)x( 1-√(1-(4.6xMu/Fck b.d.d)))
- fck = strenth of concrete
- fy = 415 N/Sq.mm
- Mu = Sp 16 P 10 Table C =0.138 . fck . b.d.d= My
Check for Cracking
Along Lx:
3d ——— Where. d = Effective depth
300mm
Spacing should not exceed smaller these two values.
Along Ly:
3d ——— Where. d = Effective depth
300mm
Spacing should not exceed smaller these two values.
Check for Deflection
Allowable L/d = 35 X M.F.X 0.8
- M.F is Obtained from IS:456-200 P-38 Fig 4
Find actual, L/d
If Actual L/d < allowable L/d ———- Ok
Check for Development Length
IS 456-2000,P.44, Cl. 26.2.3.3 C
Ld should be ≤ 1.3 (M1/V) + L0
Where
Ld = (Ø.σs / 4 τ bd )—————–σs = 0.87 fy As per IS 456-200, P.42
50 % of steel is bent up near support. Therefore find M.R for 50 % of steel only
M1 = M.R. for 50% steel support
V = Shear force at the support
L0 = Sum of anchorage beyond the center of support
d 12 Ø
Take L0 as the smaller of two values.x
Two Way Simply Support Slab Calculation /Design Example
Sum Point Consider As below
Slab Size 3.0m X 3.0 m
The slab is resting on 300mm thick wall
Find One Way Slab or Two Way Slab
ly/lx = 3.0 / 3.0 = 1 > 2
As per the type of slab
Here, this Two-way slab, So we design the slab as Two-way simply supported slab
1. Effective Depth (d)
Length – 3 m
l/d = ( 35 x M.F.) x 0.8 | IS 456-2000, P-39
fy – 415 N/sq.mm, fs = 240 N / Sq.mm
M.F = 1.3 | As per IS Code 456, Fig.4
3000 / d = (35 x 1.5 x 0.8)
d = 71.72 mm
Here, d = 100 mm , Assume 10 mm Ø bars
Overall Depth, D = 100 + (Ø / 2 ) + Clear Cover
D = 125 mm
2. Effective Span
CLear Span + d = 3000 + 100 =3100
C/c of Supports = 3000 + 230 = 3230 mm
Takeing Smaller of two values
lx , ly = 3100 mm
3. Load Calculations
| Dead load | 3.125 | Kn/m |
| Floor Finsh | 1 | Kn/m |
| Live Load | 2.5 | Kn/m |
| Total Load | 6.625 | Kn/m |
Factored Load = 1.5 x 6.625
w = 9.94 kn/sq.mm.
4. Mid Span Moment
Here, Corners down condition is given | IS 456-2000, P-90, C.L.- D1
Mx = ∝x . w. lx2
My = ∝y . w. ly2
Ly/Lx = 3100 / 3100 =1.0
∝x = 0.056
∝y = 0.056 | IS 456-2000, P-91, Table -26, case -9
Mx = ∝x . w. lx2
Mx = 0.056 x 9.94 x 3.12
Mx = My = 5.35 kn/m.
5. Effective Depth of Flexure
Mu = 0.138 fck bd2 | Sp.16, P-10 Table-c
5.35 x 106 = 0.138 x 20 x 1000 x d2
d = 44.27 mm
d = 44.27 mm < d = 100 mm …………….. o.k.
6. Reinforcement in Mild Strip
Along Lx
Pt = 50 (fck/fy) [1 – √ ( 1 -{(4.6 Mu)/ (fck x bd2)})]
Pt = 50 x (0.482) x (0.063)
Pt = 0.151%
Ast = ( pt / 100) x 1000 x 100
Ast = 150 mm2
For Spacing
Sapcing = ({[π/4] x d2}/Ast ) x 1000 mm
Sapcing = ({[π/4] x 82}/150 ) x 1000 mm
Sapcing = ( 50.26/ 150 ) x 1000 = 335 mm
Provide 8 Ø – 300 mm c/c
Along Ly
d = Depth – Dia = 100 – 8 = 92
Pt = 50 (fck/fy) [1 – √ ( 1 -{(4.6 Mu)/ (fck x bd2)})]
Pt = 50 x (0.482) x (0.075)
Pt = 0.180%
Ast = ( pt / 100) x 1000 x 92
Ast = 165.6mm2
For Spacing
Sapcing = ({[π/4] x d2}/Ast ) x 1000 mm
Sapcing = ({[π/4] x 82}/165.6 ) x 1000 mm
Sapcing = ( 50.26/ 165.6) x 1000 = 270 mm
Provide 8 Ø – 270 mm c/c
7. Check for Cracking
Along Lx
1). 3 d = 3 x 100 = 300 mm
2). 300 mm | IS 456-2000 P-46
300 mm provided < 300 mm …………….. o.k.
Along Ly
1). 3 d = 3 x 92 = 276 mm
2). 300 mm | IS 456-2000 P-46
270 mm provided < 300 mm …………….. o.k.
8. Check for Deflection
This check shall be done along with lx
Allowable (l/d) = 35 x M.F. x 0.8
% pt Provided = 100 Ast / bd = (100 x 300) / (1000 x 100) = 0.21% | IS Code 456-2000 P.38, Fig 4
M.F. = 1.7
Allowale 35 x M.F. x 0.8 = 35 x 1.7 x 0.8 =47.6
l/d = 3100 / 100 = 31
31 < 47.6 …………….. o.k.
9. Check for Development Length
Usually bond is critical along a long span
Ld ≤ 1.3 x ( M1 / V) +L0 | IS 456-2000 , P-44
For L0
1. d = 100 mm
2. 12 Ø = 12 x 8 = 96 mm
L0 = 100 mm
Steel is not bent up near support
Ast = 150 mm2
M1 = 0.87 x 415 x 1150 x 100 x [1-(415 x 150) / (20 x 1000 x 100)]
M1 = 5.85 x 106 N.mm
S.F. at support = Vu = w. lx /2 = ( 9.94 x 3.1 ) / 2 = 15.407 kn
1.3 x (M1/V) +L0 = 1.3 x ( 5.35 x 106 ) / (15.407 x 103 ) = 451 mm
For,
M-20, Fe-415
8 Ø Bar tension Ld = 376 mm | SP 16, P-184, Table – 65
Ld = 376mm < 451 mm …………….. o.k.
Two Way Simply Support Slab Calculation /Design Excel Sheet – Download
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Suggested Read –
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