C9: DESIGN OF PLINTH BEAM AS PER IS CODE

 PLINTH BEAM DESIGN

LONGER SPAN:

Span,                             L  = 5.248 mm

          Assume overall depth, D  = 700 mm = 0.70 m

Clear cover                        = 25 mm

 Assume 16 mm dia bars

EFFECTIVE DEPTH:

 d = D – cc – ( dia / 2 ) – dia

    = 700 – 25 – ( 16 / 2 ) – 16

 d = 651 mm ≃ 0.651 m

               Breadth of beam, b = 230 mm = 0.230 m

LOAD CALCULATION:

               Load from beam,

                Dead load = 37.72 KN/m

                 Live load = 5 KN/m

               Load from lintel = 27 KN/m

Self weight of earth beam = L x b x d x γ

                                          = 1 x 0.23 x 0.651 x 25

                                          = 3.74 KN/m

                         Total load = 73.46 KN/m

                   Factored load = 1.5 x 73.46 = 110.19 KN/m

BENDING MOMENT:

Mu = ( W x l² ) / 12

       = ( 110.19 x 5.248² ) / 12

       = 252.89 KNm

CHECK FOR DEPTH:

d² = (Mu) / (0.138 x fck x b)

     = ( 252.89 x 10⁶) / (0.138 x  20 x 230)

  d = 631 mm

∴ 651 mm > 631 mm

               Hence safe.

Overall depth, D  = 631 + 25 + ( 16 / 2 ) + 16  = 680 mm ≃ 700 mm

AREA OF STEEL REQUIRED:

     Xu (max) = 0.48 x d = 0.48 x 651 = 312.48 mm

               Mu = 0.87 x fy x Ast x [ d – (0.42 x Xu max)]

249.91 x 10⁶ = 0.87 x 415 x Ast x [ 651 – ( 0.42 x 312.48 )]

               Ast = 1331.72 mm²

           Provide 16 mm dia bars.

    No of bars = Ast min / ast min

                      = 1331.72 / 201.06 = 6.339 ≃ 8 nos

         Ast min = ( 0.85 x b x d ) / fy

                       = ( 0.85 x 230 x 700 ) / 415

         Ast min = 337.52 mm² ≃ 340 mm²

CHECK FOR SHEAR:

Vu = ( W x l ) / 2 = ( 110.19 x 5.248 ) / 2 = 289 KN

          Nominal shear stress (τv) = Vu / bd

           = ( 289*10³) / (230*651)

            τv  = 1.93 N/mm²

% of tensile reinforcement Pt = 100Ast / bd

 = ( 1331.72*100 ) / ( 230*651 )

 = 0.889 %

From IS 456-2000, table 19.,

τc = 0.608 N/mm²

                                  1.01 N/mm² > 0.287 N/mm²

τc > τv

                Hence OK.

DESIGN OF SHEAR REINFORCEMENT:

Assume 10 mm bars lateral ties 2 legged stirrups.

Vus = Vu – ( Zc x b x d )

        = ( 289x10³ ) – ( 0.608 x 230 x 651 )

Vus = 197.96 KN

a)      Spacing of stirrups as per design shear,

Sv  = ( 0.87 x fy x Asv x d ) / Vus

      = ( 0.87 x 415 x 78.54 x 651 ) / ( 197.96x10³ )

 Sv = 93.2 mm ≃ 90 m

b)      Spacing as per min shear reinforcement,

Sv = ( 2 x Asv x fy ) / ( 0.4 x b )

     = ( 2 x 78.54 x 415 ) / ( 0.4 x 230 )

Sv = 708 mm

c)      Spacing should not exceed,

     = 0.75 x d = 0.75 x 651 = 488.25 mm

d)     Spacing should be also not exceed,

        = 450 mm

Provide 10 mm dia bars of 2 legged stirrups @ 90 mm c/c spacing

RESULT:

Breadth,              b = 230 mm

Effective depth, d = 651 mm

Overall depth,    D = 700 mm

Total load,         W = 108.89 KN/m

     Provide 8 nos of 16 mm dia bars

     Provide 10 mm dia bars of 2 legged stirrups @ 90 mm c/c spacing.

SHORTER SPAN:

 Span,                            L  = 3.280 mm

          Assume overall depth, D  = 700 mm = 0.70 m

Clear cover                        = 25 mm

Assume 16 mm dia bars

EFFECTIVE DEPTH:

 d = D – cc – ( dia / 2 )

    = 700 – 25 – ( 16 / 2 )

 d = 667 mm ≃ 0.670 m

                  Breadth of beam, b = 230 mm = 0.230 m

LOAD CALCULATION:

               Load from beam,

               Dead load = 37.72 KN/m

                Live load = 5 KN/m

               Load from lintel = 17.07 KN/m

Self weight of earth beam = L x b x d x γ

                                          = 1 x 0.23 x 0.670 x 25

                                          = 3.85 KN/m

                         Total load = 63.64 KN/m

                    Factored load = 1.5 x 63.64 = 95.46 KN/m

BENDING MOMENT:

Mu = ( W x l² ) / 12

       = ( 95.46 x 5.248² ) / 12

       = 85.58 KNm

CHECK FOR DEPTH:

d² = (Mu) / (0.138 x fck x b)

     = ( 85.58 x 10⁶) / (0.138 x  20 x 230)

  d = 367 mm

              ∴ 670 mm > 367 mm

               Hence safe.

Overall depth, D  = 631 + 25 + ( 16 / 2 )  = 400 mm.

AREA OF STEEL REQUIRED:

   Xu (max) = 0.48 x d = 0.48 x 670 = 321.6 mm

             Mu = 0.87 x fy x Ast x [ d – (0.42 x Xu max)]

85.05 x 10⁶ = 0.87 x 415 x Ast x [ 70 – ( 0.42 x 321.6 )]

             Ast = 443 mm²

         Provide 16 mm dia bars.

 No of bars = Ast min / ast min

                   = 443 / 201.06

                        = 2.20 ≃ 4 nos

      Ast min = ( 0.87 x b x d ) / fy

                    = ( 0.85 x 230 x 670 ) / 415

      Ast min = 323.05 mm² ≃ 325 mm²

CHECK FOR SHEAR:

Vu = ( W x l ) / 2 = ( 95.46 x 3.280 ) / 2 = 156.55 KN

          Nominal shear stress (τv) = Vu / bd

           = ( 156.55*10³) / (230*670)

            τv  = 1.01 N/mm²

% of tensile reinforcement Pt = 100Ast / bd

 = ( 443*100 ) / ( 230*670 )

 = 0.287 %

        From IS 456-2000, table 19.,

                      τc = 0.48 N/mm²

                                             1.01 N/mm² > 0.287 N/mm²

                  τc > τv

                                  Hence OK.

DESIGN OF SHEAR REINFORCEMENT:

Assume 10 mm bars lateral ties 2 legged stirrups.

Vus = Vu – ( Zc x b x d )

        = ( 156.55x10³ ) – ( 0.48 x 230 x 670 )

Vus = 82.582 KN

a)      Spacing of stirrups as per design shear,

Sv  = ( 0.87 x fy x Asv x d ) /  Vus

      = ( 0.87 x 415 x 78.54 x 700 ) / ( 82.582x10³ )

Sv = 240.36 mm ≃ 240 mm

b)      Spacing as per min shear reinforcement,

Sv = ( 2 x Asv x fy ) / ( 0.4 x b )

     = ( 2 x 78.54 x 415 ) / ( 0.4 x 230 )

Sv = 708.56 mm ≃ 710 mm

c)      Spacing should not exceed,

     = 0.75 x d = 0.75 x 670 = 502.5 mm

d)     Spacing should be also not exceed,

         = 450 mm

Provide 10 mm dia bars of 2 legged stirrups @ 240 mm c/c spacing

RESULT:

Breadth,              b  = 230 mm

Effective depth, d   = 651 mm

Overall depth,    D  = 700 mm

Total load,         W  = 95.46  KN/m

     Provide 4 nos of 16 mm dia bars

                        Provide 10 mm dia bars of 2 legged stirrups @ 240 mm c/c spacing.