# 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

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. lx ^{2}**

**My = ∝y . w. ly ^{2}**

Ly/Lx = 3100 / 3100 =1.0

**∝x = **0.056

**∝y = **0.056 | IS 456-2000, P-91, Table -26, case -9

**Mx = ∝x . w. lx ^{2}**

Mx = 0.056 x 9.94 x 3.1^{2}

Mx = My = 5.35 kn/m.

**5. Effective Depth of Flexure**

**Mu = 0.138 fck bd ^{2}** | Sp.16, P-10 Table-c

5.35 x 10** ^{6 }**= 0.138 x 20 x 1000 x d

^{2}

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 bd^{2})})]

Pt = 50 x (0.482) x (0.063)

Pt = 0.151%

Ast = ( pt / 100) x 1000 x 100

Ast = 150 mm^{2}

**For Spacing **

Sapcing = ({[π/4] x d^{2}}/Ast ) x 1000 mm

Sapcing = ({[π/4] x 8^{2}}/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 bd^{2})})]

Pt = 50 x (0.482) x (0.075)

Pt = 0.180%

Ast = ( pt / 100) x 1000 x 92

Ast = 165.6mm^{2}

**For Spacing **

Sapcing = ({[π/4] x d^{2}}/Ast ) x 1000 mm

Sapcing = ({[π/4] x 8^{2}}/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 mm^{2}

M1 = 0.87 x 415 x 1150 x 100 x [1-(415 x 150) / (20 x 1000 x 100)]

M1 = 5.85 x 10^{6} 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 10^{6} ) / (15.407 x 10^{3} ) = 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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Also, read: How to Structural Design a Building/House Step by Step Part-1 (One Way Simply Support Slab)

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