# Cantilever beam

## Design calculation for cantilever beam

**14.76 kN**

**0**

**12.64 kN/m**( 230mm thick wall )

**1.72 kN/m**

**14.36 kN/m**

**15.85 kN⋅m**

**M**

_{u}= 23.77 kN⋅m

M_{ulim} = 0.138 f_{ck} ⋅bd^{2}

^{2}

^{}

**46.27**

**kN⋅m**

**(**

**M**

_{u }**<**M

_{ulim}

**...... Under reinforced section )**

_{t },

**p**

_{t = 0.513 %}_{}

_{To find the area of steel,}

_{}

**`A_{st}` = 318.57 mm**

^{2}**3 nos of 12 mm bars.**

## Point load on the simply supported beam

Point load on B-4 |

**7.125 kN/m**

**6.125**

**kN/m**

**4.06**

**kN/m**

**= 5.54 kN/m**

**9.60 kN/m**

**11.95 kN/m**

**2.58 kN/m**

**24.13 kN/m**

**143.91 kN⋅m**

**21.64 kN**

**25.17 kN⋅m**

**37.76**

**kN⋅m**

**181.67**

**kN⋅m**

_{u}= x

_{umax}

_{}

_{t}= p

_{tlim}

_{}

_{u}= M

_{ulim}

**d = 535.45 mm**

**D = 600 mm**if we design as a singly reinforced beam.

_{u}= 181.67 kN/m

_{ulim}= 0.138⋅f

_{ck}⋅b⋅d

^{2}(SP. - 16, P- 10, table - C )

_{ulim}for 9 X 18 = 111.97 kN/m < 181.67 kN/m (d = 420)

### The beam size of 9 X 24-inch

### Total slab weight = 9.60 kN/m

### Wall weight = 0.230 X (3.04 - .600) X 20

### = 11.22 kN/m

### Self-weigth of beam = 0.230 X 0.600 X 25

### = 3.45 kN/m

**24.27 kN/m**

^{2}/ 8

**144.75 kN⋅m**

**182.50**

**kN⋅m**

_{ }`p_{t} = 50 \frac{f_{ck}}{f_{y}} [1 - \sqrt{1-\frac{4.6M_{u}}{f_{ck} b d^{2}}} ]`

**p**

_{t}= 0.814 %

**A _{st = 1067.15 }mm^{2 }**

^{}

^{Provide, 1067.15 / 201}

^{5 nos 16 mm bars.}

^{}

^{If we design as a singly reinforced beam area of steel is 1067.15 }^{mm2. }

^{}

^{what if the size is limited to 9 X 21 for B-4?}

^{}

^{For that, we can design a doubly reinforced beam section.}

^{}

^{Doubly reinforced beam}

^{Self-weight of beam = 0.230 X 0.530 X 25}

^{}

^{ = 3.04 kN/m}

^{}

^{Wall weight of beam = 0.230 X (3.04 - 0.530) X 20}

^{}

^{ = 11.54 kN/m}

^{Slab weight = 9.60 + 3.04 + 11.54 }

^{}

^{ = 24.18 kN/m}

^{Factored moment due to Udl = 144.22 kN⋅m}

^{}

^{Total factored moment }M_{u} = 144.22 + 37.755

= **181.98 kN⋅m**

M_{ulim = 158.7 kN⋅m}

_{}

M_{u2 = }M_{u - }M_{ulim}

_{}

M_{u2 = 181.98 - 158.7}

_{ = 23.28 kN⋅m}

_{}

f_{cc = 0.446 }f_{ck ( SP. 16, P-13)}

_{}

_{ = 0.446 X 20}

_{}

_{ = 8.92}

_{}

_{}

d'/d = 30/500 ( SP. 16, P-13, Table - F)

= 0.06 ≈ 0.10

f_{sc = 353}_{ N/mm}2

^{To Find}^{ A}sc^{ (IS: 456-2000, P- 96 , CL. G-1.2)}

$$M_{u2} = (f_{sc} - f_{cc}) A_{sc} (d-d')$$

`23.28 X 10^{6} = (353 - 8.92) A_{sc} (500-30)`

**`A_{sc} = 143.95 mm^{2}`**

Find A_{st1, }

_{`p_{tlim} = 19.82 X (\frac{f_{ck}}{f_{y}})`}

_{}

_{`p_{tlim}` = 0.96 %}

_{}

`\therefore A_{st1} = \frac{p_{tlim}}{100} X bd`

** `A_{st1}` = 1104 mm ^{2}**

^{}

^{Find}^{ A}st2 (IS: 456-2000, p-96),

$$A_{st2} = \frac{(f_{sc} - f_{cc}) A_{sc}}{0.87 f_{y}}$$

$$= \frac{(353- 8.92) X 143.95}{0.87 X 415} = 137.18 mm^{2}$$

`A_{st} = A_{st1} + A_{st2}`

**`A_{st}` = 1241.18 mm ^{2}**

Provide reinforcements,

Top = 2 nos. 12 mm dia. bars

Bottom = 2 nos. 20 mm dia. and 3 nos. 16 mm dia. bars.

#### Summary

^{2 }and 9 X 21 requires 1241.18 mm

^{2}area of steel.

**Revise**the beam layout plan and try to reduce the length of the beam and avoid point load if possible.

**partition wall**is possible on the first-floor balcony.

**factor of safety**for concrete = 1.5 which raises the moment on the beam. Instead of 1.5, we can even use 1.4,1.3 or 1.2 according to some other standard code.

**design software**. try to use the software such as ETABS, Staad pro, struds, SAP 2000, etc.

## Conclusion

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