C7: DESIGN OF LINTEL AS PER IS CODE

DESIGN OF LINTEL

LONGER SPAN:

Size of  lintel = 230 mm x 230 mm

Clear cover    = 20 mm

fck = 20 N/mm ² & fy = 415 N/mm²

Use 8 mm dia bars.

EFFECTIVE DEPTH:

       d = D – cc – ( dia / 2) – dia

          = 230 – 20 – ( 8 / 2) – 8

          = 198 mm

                   Clear width ,                                    L  = 5.248 m

Height of masonry above the opening = 0.9 m

Height of masonry above the lintel      = 0.9 - 0.16 = 0.74 m

Assume the weight of masonry wall    = 15 KN/m³

LOAD CALCULATION:

Weight of masonry        = 5.248 x 0.74 x 0.23 x 15 = 13.39 KN

Self weight of masonry = 5.248 x 0.16 x 0.23 x 25 = 4.828 KN

Total load                      = 13.39 + 4.828 = 18.218 KN

Design load                   = 1.5 x 18.218 = 27.327 KN

BENDING MOMENT:

BM, Mu = ( W x L ) / 8

               = ( 27.327 x 5.248 ) / 8

               = 17.92 KNm.

CHECK FOR EFFECTIVE DEPTH:

      d² = (Mu) / (0.138 x fck x b)

          = ( 17.92 x 10⁶) / (0.138 x  20 x 230)

       d =  168 mm

                     ∴ 198 mm > 168 mm

               Hence safe.

Overall depth = 168 + 20 + ( 8 / 2) + 8 = 200 mm.

AREA OF STEEL:

                           Mu = 0.87*fy*Ast*d*[(1)-((Ast*fy) / (fck*b*d))]

                17.92*10⁶ = 0.87*415*Ast*198*[(1)-((Ast*415) / (20*230*198))]

Ast = 288 mm²

MINIMUM AREA OF STEEL:

    Ast min = ( 0.85 x b x d ) / fy

                  = ( 0.85 x 230 x 198 ) / 415

    Ast min = 93.27 mm²

No of bars = Ast min / ast  = 288 / 93.27  = 3.08 ≃ 4 nos

∴ provide 4 nos of 10 mm dia bars

Using 6 mm dia of 2 legged stirrups

i ) spacuing = ( Asv x fy ) / ( 0.4 x b )

                    = ( 28.27 x 415) / ( 0.4 x 230 )

                    = 127.52 mm ≃ 130 mm

ii ) 0.75 x d  = 0.75 x 198 = 148.5 mm ≃ 150 mm

iii ) 450 mm.

RESULT:

Size of lintel          = 230 mm x 230 mm

Overall depth,    D = 200 mm

Effective  depth, d = 198 mm

Provide 4 nos dia of  bars

     Provide 6 nos dia of lateral ties @ 150 mm c/c spacing

SHORTER SPAN:

Size of  lintel = 230 mm x 230 mm

Clear cover    = 20 mm

fck = 20 N/mm 2 & fy = 415 N/mm²

Use 8 mm dia bars.

EFFECTIVE DEPTH:

          d = D – cc – ( dia / 2)

             = 230 – 20 – ( 8 / 2)

             = 206 mm

Clear width span,                              L = 3.280 m

Height of masonry above the opening = 0.9 m

Height of masonry above the lintel      = 0.9 - 0.16 = 0.74 m

Assume the weight of masonry wall    = 15 KN/m³

LOAD CALCULATION:

Weight of masonry       = 3.280 x 0.74 x 0.23 x 15 = 8.37 KN

Self weight of masonry = 3.280 x 0.16 x 0.23 x 25 = 3.0176 KN

Total load                      = 8.37 + 3.0176 = 11.38 KN

Design load                   = 1.5 x 11.38 = 17.07 KN

BENDING MOMENT:

BM, Mu = ( W x L ) / 8

               = ( 17.07 x 3.280 ) / 8

               = 6.9987 KNm.

CHECK FOR EFFECTIVE DEPTH:

       d² = (Mu) / (0.138 x fck x b)

           = ( 6.99 x 10⁶) / (0.138 x  20 x 230)

        d =  63 mm ≃ 70 mm

    ∴ 206 mm > 70 mm

               Hence safe.

Overall depth = 70 + 20 + ( 8 / 2) = 94 mm ≃ 100 mm.

AREA OF STEEL:

                            Mu = 0.87*fy*Ast*d*[(1)-((Ast*fy) / (fck*b*d))]

                   6.99*10⁶ = 0.87*415*Ast*206*[ (1)-((Ast*415) / (20*230*206 ))]

Ast = 98 mm2

MINIMUM AREA OF STEEL:

   Aast min = ( 0.85 x b x d ) / fy

                   = ( 0.85 x 230 x 206 ) / 415

     Ast min = 97.04 mm2

 No of bars = Ast min / ast min

                   = 98 / 97.04

                   = 1.009 ≃ 2 nos

∴ provide 2 nos of 10 mm dia bars

Using 6 mm dia of 2 legged stirrups

i ) spacing = ( Asv x fy ) / ( 0.4 x b )

                    = ( 2 x 28.27 x 415) / ( 0.4 x 230 )

                    = 254.9 mm ≃ 260 mm

ii ) 0.75 x d = 0.75 x 206 = 154.5 mm ≃ 160 mm

iii ) 450 mm.

RESULT:

Size of lintel            = 230 mm x 230 mm

Overall depth,     D  = 206 mm

Effective  depth,  d  = 100 mm

Provide 2 nos dia of  bars

Provide 6 nos dia of lateral ties @ 160 mm c/c spacing