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How to Structural Design a Building/House Step by Step Part-2 (Two Way Simply Support Slab)

Important Point

    • Two Way Simply Support Slab Calculation /Design
    • Effective Depth (d)
    • Effective Span
    • Load Calculations
    • Mid Span Moment
    • Effective Depth of Flexure
    • Reinforcement in  Mild Strip
    • Check for Cracking
    • Check for Deflection
    • Check for Development Length
    • Two Way Simply Support Slab Calculation /Design Example
    • Two Way Simply Support Slab Calculation /Design Excel Sheet – Download 
  • Video tutorial for better understanding:

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

Two Way Slab

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 Slab

Two Way Simply Support Slab Calculation /Design Excel Sheet – Download 

Video tutorial for better understanding:

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Suggested Read –

Also, read: How to Structural Design a Building/House Step by Step Part-1 (One Way Simply Support Slab)

Also, read: How to Structural Design a Building/House Step by Step Part-3 (Slab Beam Design)

Also, read: How to Structural Design a Building/House Step by Step Part-4 (Column Design)

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