Steel Reinforcement Fe415 and Fe500 with Bar Bending Schedule Basics

Concrete is strong in compression but weak in tension, so reinforced concrete uses steel bars to carry tensile forces. On an Indian site, the junior engineer is expected to read the structural drawing, order the right bars and prepare a bar bending schedule (BBS) that tells the bar benders exactly how to cut and bend each bar. This article covers steel grades, unit weights, development length and a complete BBS for a simple beam.

Reinforcement Grades

High strength deformed bars, sold as TMT (thermo-mechanically treated) bars, are covered by IS 1786. The grade number is the minimum 0.2 percent proof stress or yield stress, fy, in N/mm². Grades with the suffix D, such as Fe500D, have tighter chemistry limits and higher elongation, giving better ductility for earthquake-resistant construction.

Grade	fy (N/mm²)	Min elongation (%)	Remarks
Fe415	415	14.5	Older standard grade, still common in design examples
Fe500	500	12.0	Most widely used grade today
Fe500D	500	16.0	Preferred in seismic zones for ductility
Fe550D	550	14.5	Heavy structures, high-rise columns

In limit state design per IS 456, the design stress of steel is 0.87 fy, so Fe415 gives 361 N/mm² and Fe500 gives 435 N/mm².

Unit Weight of Bars

Steel is bought by weight but detailed by length. The density of steel is about 7850 kg/m³, which leads to a convenient formula.

Weight per metre (kg/m) = d² / 162     (d in mm)

Derivation: (π/4) x d² x 1 m x 7850 kg/m³ / 10^6 = d² / 162.2

Diameter (mm)	8	10	12	16	20	25
kg per metre	0.395	0.617	0.889	1.580	2.469	3.858
Area (mm²)	50.3	78.5	113.1	201.1	314.2	490.9

Development Length

A bar must be embedded long enough so that bond between steel and concrete can develop its full design stress. IS 456 gives the development length as follows.

Ld = (φ x σs) / (4 x τbd)

φ   = bar diameter
σs  = stress in bar = 0.87 fy at full design stress
τbd = design bond stress; for M20 plain bars = 1.2 N/mm²,
      increased by 60 % for deformed bars = 1.92 N/mm²

Fe415 bar in M20:
Ld = φ x 0.87 x 415 / (4 x 1.92) = φ x 361.05 / 7.68 = 47 φ

For a 16 mm bar: Ld = 47 x 16 = 752 mm

This is why you will often hear site engineers say 47d or 50d for laps and anchorage in M20. The figure changes with concrete grade and steel grade, so always compute it rather than memorising a single number.

Cutting Length and Bend Deductions

When a bar is bent, the outer face stretches, so the cutting length is slightly less than the sum of the outside dimensions. Sites use simple deduction rules for this. The values below are widely used site conventions; your project specification or detailing standard may use slightly different figures.

Bend angle	Deduction
45°	1d
90°	2d
135°	3d
180°	4d

Stirrup cutting length

Take a beam 230 mm wide and 450 mm deep with 25 mm clear cover and 8 mm two-legged stirrups having 135° hooks of 10d at each end.

Stirrup dimension a = 230 - 2 x 25 = 180 mm
Stirrup dimension b = 450 - 2 x 25 = 400 mm

Cutting length = 2(a + b) + 2 x hook - bend deductions
               = 2(180 + 400) + 2 x (10 x 8) - [3 x (2 x 8) + 2 x (3 x 8)]
               = 1160 + 160 - (48 + 48)
               = 1224 mm

Worked Example: BBS for an RCC Beam

Beam B1 is 4.0 m long overall, 230 x 450 mm, with 25 mm cover. Reinforcement: 3 bars of 16 mm at the bottom with 150 mm L-bends at both ends, 2 bars of 12 mm straight at the top, and 8 mm stirrups at 150 mm centres with the first stirrup 50 mm from each end.

Bottom bars (16 mm):
  Straight length = 4000 - 2 x 25 = 3950 mm
  Cutting length  = 3950 + 2 x 150 - 2 x (2 x 16) = 4186 mm

Top bars (12 mm):
  Cutting length  = 4000 - 2 x 25 = 3950 mm

Stirrups (8 mm):
  Number = (4000 - 2 x 50) / 150 + 1 = 26 + 1 = 27
  Cutting length = 1224 mm (from above)

Bar mark	Description	Dia (mm)	No.	Cutting length (m)	Total length (m)	kg/m	Weight (kg)
a	Bottom main bars, L-bent	16	3	4.186	12.558	1.580	19.84
b	Top hanger bars	12	2	3.950	7.900	0.889	7.02
c	Stirrups, 2-legged	8	27	1.224	33.048	0.395	13.05
	Total						39.91

Add about 3 percent for wastage and rolling margin when ordering: 39.91 x 1.03 is about 41 kg for this beam.

Common Mistakes

Mixing Fe415 and Fe500 bars on the same site without clear marking; the grade is rolled onto the bar, so check it at delivery.
Forgetting bend deductions, which makes stirrups too large to fit inside the cover.
Placing laps of all bars at the same section or in zones of maximum moment.
Using 90° hooks on stirrups in seismic zones where 135° hooks are required by ductile detailing provisions.

Summary

TMT bars are graded by yield stress, with Fe500 and Fe500D now standard. Weight per metre is d²/162, development length comes from Ld = φσs/4τbd, and cutting lengths account for hooks and bend deductions. A BBS table lists each bar mark, diameter, number, cutting length and weight, and is the basis for ordering steel and checking bills.