One Way Slab and Two Way Slab Design Step by Step

One Way Slab and Two Way Slab Design Step by Step (1)

Design Considerations: 

One Way Slab

One way slab

1. Effective Depth (d)

For deflection control 

L/d = 20 X M.F

  • 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.

2. Effective Span:

ClearSpan + d

c/c of Supports

Whichever is smaller ——–  as per IS 456-2000 P. 34, CI 22.2.a

3. Reinforcement Requirements

Minimum reinforcement 

For Fe-250     Pt = 0.15 % of total C/s area (d x D)

For Fe-415     Pt = 0.12 % of total C/s area (d x D)

For Fe-500     Pt = 0.12 % of total C/s area (d x D) ——–  as per IS 456-2000 P. 48, CI 26.5.2.1

Maximum diameter (Sp 34)

For minbar:

  • Plain bars———–10 mm Ø min dia
    Deformed bars—–8 mm Ø min dia

For Distribution bars:

  • Plain bars———–6 mm Ø min dia
    Deformed bars—–6 mm Ø min dia

4. Check for Cracking

For Min Steel:

3d ——— Where. d = Effective depth

300mm

Spacing should not exceed smaller these two values.

For Distribution steel:

5 d

450mm

Spacing should not exceed smaller these two values. ——- IS: 456-2000, P.46

5. Check for Deflection:

Allowable L/d  = 20 X M.F.

  • M.F is Obtained from IS:456-200 P-38 Fig 4

Find actual, L/d

If Actual L/d < allowable  L/d ———- Ok

6. Check for Development Length (Ld)

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 centre of support

 d

12  Ø

Take L0 as the smaller of two values.

Also, read: What is Chain Surveying (Principal, Procedure, Method, Instrument)

Two Way Slab

Two Way Slab

1. 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.

2. Effective Span:

  • ClearSpan + d
  • c/c of Supports

Whichever is smaller ——–  as per IS 456-2000 P. 34, CI 22.2.a

3. 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

4. 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 coefficient are obtained from IS 456 tables -26, fig 10.3 shoe nine separate possible arrangement for a two-way restrained slab.

5. 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

6. 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

7. Check for Cracking:

Along with Lx:

3d ——— Where. d = Effective depth

300mm

Spacing should not exceed smaller these two values.

Along with Ly:

3d ——— Where. d = Effective depth

300mm

Spacing should not exceed smaller these two values.

8. 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

9. 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 centre of support

 d

12  Ø

Take L0 as the smaller of two values.

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