# Design a rectangular beam over 4 column

**Design a rectangular beam continuous over 4 column supports of effective shap 6m. The beam is subjected to an imposed dead load of 10 KN/m and a live load of 15 KN/m. use M20 and Fe415 steel.**

**Given**

Effective shap (L eff. ) = 6m

(f ck ) = 20 N/mm 2

(f y ) = 415 N/mm 2

Imposed dead load = 10 KN/m

Imposed live load = 15 KN/m

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a.

**Effective depth**:-Assuming effective depth = = = 400 mm

Trying a total depth of D = 500 mm

And width (b) = to i.e,

Nearly 250mm

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Trying a width (b) of 300mm

Assuming an effective cover of 30mm

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⸫ effective depth (d) = 500 – 30 = 470mm

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**⸫ the dimensions are:-**

Width(b) = 300mm

Effective depth (d) = 470mm

Total depth (D) = 500mm

Effective depth (d) = 470mm

Total depth (D) = 500mm

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**b. Load calculation:-**

Self weight of beam = 0.3×0.5×25 = 3.75 KN/m

Imposed dead load = 10 KN/m

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⸫ total dead load = 10 + 3.75 = 13.75 KN/m

⸫ factored dead load (W d )= 13.75 x 1.5 = 20.625 KN/m

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⸫ factored live load (W L ) = 15 x 1.5 = 22.5 KN/m

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**c. B.M. & S.F. using coefficient:-**

(ref. table- 12,13, IS 456:2000)

Now,

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M = WL 2

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the maximum negative moment occurs at support next to end support,

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M = – (20.625 x 6 2 ) + (22.5 x 6 2 )

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= -164.25 KNm

The maximum positive moment occurs near the middle of the end shap,

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M = (20.625 x 6 2 ) + (22.5 x 6 2 )

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= 142.875 KNm

And,

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V = WL

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Maximum shear force occure at outer side of the support next to end support,

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V u = 0.6 ( 20.625×6 ) + 0.6 ( 22.5×6 )

= 155.25 KN

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**d. Depth check :-**

( Ref. CL. 31.1.1, IS 456 : 2000 )

M u = 0.36 bd 2 f ck 1- 0.42

= 0.36 x 0.48 x 1- ( 0.42 x 0.48 )

= 0.138 f ck bd 2

d req. =

=

= 446 mm < 470 mm

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**e. Area of steel :-**

(Ref. CL. 31.1.1 IS 456 : 2000 )

Calculating the limiting moment of resisitance,

We know,

M u = 0.138 f ck bd 2

= 0.138 x 20 x 300 x 470 2

= 182.9 KNm > M u (164.28KNm)

Hence, the singly reinforced section can be designed,

Ast = 0.5 1 –

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Now,

Designing the section at intermediate support,

M u = 0.87 f y Ast d 1-

164.25 x 10 6 = 0.87 x 415 x Ast x 470 1 –

Solving we get,

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Ast = 1014 mm 2

No o bars = 4

Using 4 – 20 mm ϕ bars (Ast Pro. = 1256 mm 2 ) at intermediate support as negative

Reinforcement.

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**At mid-span –**

M u = 142.875 KNm

M u = 0.87 f y Ast d 1-

142.875 x 10 6 = 0.87 x 415 x Ast x 470 1-

Solving we get,

Ast = 876 mm 2

No. of bars = 3

Using 3 – 20 mm ϕ bars (Ast = 942 mm 2 ) at mid-span as positive Reinforcement.

Ast = (Ref. CL 26.5.1.1, IS 456:2000)

Ast = = 289mm 2 > Ast

Hence, Ast provided is OK.

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**f. Shear design :-**

At support next to end support,

V u = 155.25KN

τ v = (Ref. CL 40.1 IS 456:2000)

τ v = = 1.1 N/mm 2

from table 20, IS 456:2000, for M20,

τ c = 2.8 N/mm 2 > τ v

At support,

P t = 100 x = = 0.89%

From table 19, IS 456:2000, For M20, concrete & P t = 0.89%

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P t = 0.89 N/mm 2 < τ v

Hence, shear reinforcement is necessary

V us = V o – τ c bd (Ref. of CL 40.4)

V us = 155.25 x 10 3 – 0.02 x 300 x 470 = 67830 N

Spacing of 2 legged 8mm ϕ bars stirrups

S v = (Ref. CL 40.4(a) IS 456:2000)

S v = = 251 mm

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Maximum spacing of 2 legged 8 mm dia stirrups should not excced

I. 0.75d = 0.75 x 470 = 352 mm

II. 300 mm

III. S v max =

= = 302 mm

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Hence provide 2 legged 8mm ϕ @ 250 mm c/c at end support,

V u = ( 0.4 x 20.625 x 6) +( 0.45 x 22.5 x 6 ) = 110.25 KN

τ v = = = 0.78 N/mm 2

From table 20 , IS 456 : 2000;

τ c max = 2.8 N/mm 2 > τ v

P t at end support = 100 x = = 0.66 %

From table 19, IS 456 2000,For M20 concrete,

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P t = 0.66%

τ c = 0.48 + (0.66 – 0.5 ) = 0.53 N/mm 2 < τ v

Hence, shear reinforcement is necessary

V us = V u – τ c bd (Ref. CL 40.4 IS 456:2000 )

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V us = (110.25 x 10 3 ) – (0.53 x 300 x 470 )

V us = 35520 N

Spacing of 2 legged 8 mm ϕ stirrups

S v = (Ref. CL 40.4(a) IS 456:2000)

S v = = 480 mm > 300mm

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Provide 2 legged 8 mm ϕ stirrups @ 300 mm c/c. hence, provide 2 legged 8 mm ϕ stirrups @

250 mm c/c at intermediate support and gradually increasing the spacing to 300 mm c/c near the

mid-span & at the end support.