Important Point

**Design Considerations: **

### 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

#### 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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