Two Way Simply Support Slab Calculation /Design
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.
FAQ: Design and Calculation of Two-Way Simply Supported Slab
What is a two-way simply supported slab?
A two-way simply supported slab is a slab that is supported on all four sides, with the supports allowing rotation but preventing translation. It distributes loads in two directions, making it effective for square or nearly square layouts.
How do you determine if a slab is one-way or two-way?
To determine if a slab is one-way or two-way, calculate the aspect ratio (ly/lx). If the ratio is less than 2, the slab is considered a two-way slab; otherwise, it is a one-way slab.
What is the effective depth (d) and how is it calculated?
The effective depth (d) is the distance from the top of the slab to the centroid of the reinforcement. It is calculated based on the span-to-depth ratio, modified by factors such as the type of reinforcement and concrete strength, as per IS 456-2000 guidelines.
How is the effective span of the slab determined?
The effective span is the smaller of the clear span plus the effective depth (d) and the center-to-center distance of supports. This is as per IS 456-2000, Clause 22.2.a.
What loads are considered in the load calculation?
The load calculation includes the dead load of the slab, floor finishing load, live load, and any other applied loads. The total load is then factored by a safety factor, typically 1.5, to get the factored load.
How is the mid-span moment calculated for a two-way slab?
The mid-span moment is calculated using coefficients for moments (αx and αy) from IS 456-2000, which depend on the aspect ratio and boundary conditions. The formula is Mx = αx * w * lx^2 and My = αy * w * ly^2.
How do you determine the reinforcement required in the mild strip?
The required reinforcement is calculated using the percentage of tensile steel (Pt), which depends on the moment of resistance (Mu), concrete strength (fck), and yield strength of steel (fy). The area of steel (Ast) is then determined, and spacing of bars is calculated accordingly.
What checks are performed to ensure the slab design is safe?
Several checks are performed, including:
- Cracking Check: Ensuring the spacing of reinforcement bars is within the limits to control cracks.
- Deflection Check: Comparing the actual span-to-depth ratio with the allowable ratio to ensure the slab does not deflect excessively.
- Development Length Check: Ensuring the provided development length for reinforcement bars is adequate for bond strength.
How is the development length (Ld) calculated?
The development length is calculated using the formula Ld = (Ø * σs) / (4 * τbd), where Ø is the diameter of the bar, σs is the stress in steel, and τbd is the design bond stress. The result is checked against the requirement of Ld ≤ 1.3 * (M1/V) + L0 as per IS 456-2000.
What standards and codes are referred to in the slab design process?
The slab design process primarily refers to IS 456-2000, which provides the guidelines for the design of reinforced concrete structures in India, including tables, figures, and clauses specific to various aspects of slab design.
