Answer / bilal ali khan

development length mostly used when beam bar are passing through column and are simply shown as ld+10d
for fe500 ld is
m40-36d
m35-40d
m30-46d
m25-49d
m20-57d
for fe415 ld is
m40-30d
m35-30d
m30-38d
m25-41d
m20-47d

Answer / rajarshi basu

As per IS 456 : 2000, Page no 42 (Clause no. 26.2.1) development length Ld is given by

Ld = (s)/(4*bd)

Let nominal diameter of the bar = 16 mm

s = Stress in bar at the section considered at design load = 0.87* fy = 0.87*500 = 435 MPa

bd = design bond stress = 1.9 MPa for M40 concrete (As per table 26.2.1.1 of IS 456 : 2000)

Also just below the table 26.2.1.1,it is mentioned that
For deformed bars the value of bd shall be increased by 60%. Hence the value of bd shall be 1.9*1.6=3.04 Mpa.

So development length of tension bars of 16mm dia (Fe 500) in M40 concrete is

= (16*435)/(4*3.04) mm = 572 mm

572/16=35.75 ~36

Therefore for M40 Grade of concrete=Developement length in tension=36 x d

where,d=dia of the bar

According to IS 13920:1993, Clause 6.2.5 it is also stated that
In an external joint,both the top and the bottom bars of the beam shall be provided with anchorage length,beyond the face of the column,equal to the developement length in tension+10 x bar diameter-Allowance for 90 degree bend.(Thats means now we have to consider the Grade of the concrete,in our case it is M40)

therefore for M40 Concrete, for an external joint anchorage should be ld+10d=36d+10d=46d

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