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Tension Development Length - Decoding the K-Factors in CSA A23.3

TENSION DEVELOPMENT LENGTH above a hatched T-section with a red bar and ld dimension spanning to the right end
TL;DR Key takeaways before you dive in
  • A 25M top bar in 30 MPa concrete with minimum stirrups needs 1077 mm in 400W steel and 1501 mm in 500W under CSA A23.3-24.
  • The new grade factor \(k_g = f_{yg}/900 + 0.56\) adds 12 % on top of the 25 % from the higher yield strength, so re-check every lap and cut-off when the grade changes.
  • In a typical building only the top bar factor (1.3) and the small bar factor (0.8) move off 1.0, so check those two first.
You'll learn:
  • How the old K1 to K4 factors map to the five factors in CSA A23.3-24 Table 12.2
  • When the 0.45 multiplier applies instead of 0.6 in the simplified equation of Cl. 12.2.3
  • Why a top bar takes 1.3 and a 20M or smaller bar takes 0.8
  • How the new grade factor makes 500W bars 39 % longer to develop than 400W

Calculations referenced in this post

Development length in tension

Required Ld for bars in tension to CSA A23.3-24 Cl. 12.2

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Swapping 400W bars for 500W to save steel feels like a 25 % hit on development length, since the yield strength went up by 25 %. Under CSA A23.3-24 it’s closer to 39 %, because the 2024 edition added a fifth modification factor that grows with the bar grade. A 25M top bar in 30 MPa concrete goes from 1077 mm to 1501 mm, and that can push a cut-off or a lap splice past where the drawings put it. CSA A23.3-24 Cl. 12.2.3 and Table 12.2 set the length.

Five factors, and the two that usually move

Before any factor, the simplified equation makes you pick a multiplier, and that choice moves the answer more than most factors do. You take 0.45 when the member has minimum stirrups or ties along the whole length, or when the bars have two diameters of clear space between them and one and a half of cover. Anything else takes 0.6, which is a third longer. So a bar that comes out too long is worth checking for stirrups before you hunt through the factors, which scale the length for conditions that weaken or help bond. If you learned them as K1 to K4, the 2024 edition renamed them and added one more.

A23.3-19A23.3-24What it coversUsual value
\(k_1\)\(k_\ell\)Bar location1.3 for top bars, else 1.0
\(k_2\)\(k_c\)Coating1.0 uncoated; 1.2 or 1.5 epoxy; 1.2 galvanized
\(k_3\)\(k_d\)Concrete density1.0 normal; 1.2 semi-low; 1.3 low-density
\(k_4\)\(k_s\)Bar size0.8 for 20M and smaller; 1.0 for 25M and up
none\(k_g\)Reinforcement grade1.0 up to 400 MPa; 1.12 at 500 MPa

In most Canadian buildings the bars are uncoated and the concrete is normal density, which leaves bar location and bar size doing the real work. Bar location is the one with the surprising definition, because “top bar” is set by how much fresh concrete was placed under it, not by where the steel sits in the member. Once more than 300 mm of concrete sits below a horizontal bar, bleed water and trapped air rise and collect on its underside. So the bar ends up bonded to a weak, porous layer instead of sound paste.

Hatched rectangle with top bar kl = 1.3, bottom bar kl = 1.0, upward arrows, more than 300 mm span

That’s why the negative steel over a support in a 600 mm beam takes 1.3 while the bottom steel takes 1.0. The top mat in a 200 mm slab stays at 1.0, except where it runs over a deeper beam cast with it. Bar size goes the other way. A 20M or smaller bar has more surface for the force it carries, so it gets a 20 % reduction, and that 0.8 is one of the few factors that hands you length back.

Worked example - what 500W does to a top bar

Take four uncoated 25M top bars over an interior support of a 600 mm beam in 30 MPa normal-density concrete. The beam has minimum stirrups throughout, so the 0.45 multiplier applies, and the bars are top bars. The cut-off on the drawings sits 1200 mm past the section where the bars need their full yield stress. Switch the spec to 500W and a new factor joins in. The grade factor is \(k_g = f_{yg}/900 + 0.56\) for bars above 400 MPa, which works out to 1.12 at 500 MPa (1.1156 before rounding). The catch is that it multiplies the yield strength term, which already went up by 25 %, so the two stack to about 1.39. Bond gets less efficient as bar stress climbs, so 25 % more force needs more than 25 % more length.

Calculation

25M top bars over an interior support, 400W against 500W

CSA A23.3-24 12.2.3CSA A23.3-24 Table 12.2

Simplified equation; 400 mm web, 40 mm clear cover, 10M stirrups throughout; uncoated bars, normal-density concrete

Inputs
6 input values\(f'_c = 30\) MPa, \(d_b = 25.2\) mm, \(L_{avail} = 1200\) mm ...
\(f'_c\)30MPaSpecified concrete strength
\(d_b\)25.2mm25M bar diameter
\(L_{avail}\)1200mmLength from the critical section to the cut-off
\(k_\ell\)1.3Top bar
\(k_c k_d k_s\)1.0Uncoated
\(k_g\)1.0400W bars; 1.1156 for 500W (step 2)
Derivation 3 steps
Step 1: Development length with 400W bars\(\ell_d = 1077\ \text{mm}\)
\[\ell_d = 0.45\, k_\ell k_c k_d k_s k_g \frac{f_{yg}}{\sqrt{f'_c}} d_b\]
\[= 0.45(1.3)(1.0)(1.0)\frac{400}{\sqrt{30}}(25.2)\]
\(\ell_d = 1077\ \text{mm}\)A23.3-24 12.2.3
Step 2: Grade factor for 500W bars\(k_g = 1.1156\)
\[k_g = \frac{f_{yg}}{900} + 0.56\]
\[= \frac{500}{900} + 0.56\]
\(k_g = 1.1156\)A23.3-24 Table 12.2
Step 3: Development length with 500W bars\(\ell_d = 1501\ \text{mm}\)
\[\ell_d = 0.45\, k_\ell k_c k_d k_s k_g \frac{f_{yg}}{\sqrt{f'_c}} d_b\]
\[= 0.45(1.3)(1.0)(1.1156)\frac{500}{\sqrt{30}}(25.2)\]
\(\ell_d = 1501\ \text{mm}\)39 % longer than the 400W bar, not 25 %
Checks
Clear cover allows the simplified equationA23.3-24 12.2.3PASS
\(c_{clear} \ge d_b\) : 40 + 11.3 = 51.3 mm ≥ 25.2 mm Demand / capacity0.49
Clear spacing allows the simplified equationA23.3-24 12.2.3PASS
\(s_{clear} \ge 1.4 d_b\) : (400 - 2(40) - 2(11.3) - 4(25.2))/3 = 65.5 mm ≥ 35.3 mm Demand / capacity0.54
400W bars reach the cut-offPASS
\(\ell_d \le L_{avail}\) : 1077 mm ≤ 1200 mm Demand / capacity0.90
500W bars reach the cut-offFAIL
\(\ell_d \le L_{avail}\) : 1501 mm > 1200 mm Demand / capacity1.25
Watch outLaps for these bars are set from the same development length, so they grow by the same 39 %.
Different inputs? Development length in tension runs this check with yours. Run your own bar and grade →

Development length by bar size for 400W and 500W steel, with a 1200 mm reference line

So the 400W detail clears with about 120 mm to spare, while the same bars in 500W fall 300 mm short. You’d either extend the cut-off or drop to 20M bars, which pick up the 0.8 size factor and need only 929 mm in 500W. It takes seven to match the steel area, and seven won’t fit in one layer of this web at the spacing the equation needs. That’s the development check only, so the Cl. 11.3.9 shear extension and the Cl. 12.10 and 12.12 cut-off rules still apply. If extending isn’t possible at an exterior support, a hook is usually the next move, and the hooked bar post walks through the factors that shrink a 25M hook into a 600 mm column.

Where the simplified equation stops

The simplified equation only applies when clear cover is at least one bar diameter and clear spacing at least 1.4 diameters. The product of the location and coating factors is capped at 1.7, which only bites on top bars that take the 1.5 coating factor. No development length goes below 300 mm either, which is why a 10M bottom bar in 400W steel at 30 MPa, with the 0.45 multiplier, computes to 297 mm and still gets 300.

The excess reinforcement reduction in Cl. 12.2.5 is tempting, but it’s off the table where development of full yield is specifically required. That covers shear friction dowels, which have to develop yield on both sides of the plane, and seismic detailing under Clause 21.

The 2024 edition also reworded the spacing route to 0.45. A23.3-19 limited it to slabs, walls, shells and folded plates, while A23.3-24 opens it to any member that meets the spacing and adds a clear cover limit of 1.5 bar diameters. The catch is slabs. A 15M bar with 20 mm of cover falls under the new 24 mm limit, so without ties it drops to 0.6, and an old slab table at 0.45 comes out a quarter short.

For 400W bars in a normal building, the top bar and small bar factors still carry most of the change, as long as you recheck which multiplier the cover allows. The shortcut stops holding the moment the specification says 500W or galvanized, because each of those adds a factor that a table built on K1 to K4 never carried. That’s the point to recompute rather than look up, the same way an equation you use without seeing its assumptions turns into a design error.

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Disclaimer: This blog post is for informational purposes only and should not be taken as specific engineering advice. Always consult the latest edition of the National Building Code of Canada and relevant CSA standards for your projects.

Quick answers

Table 12.2 has five: bar location \(k_\ell\) (1.3 for top bars), coating \(k_c\) (1.2 or 1.5 for epoxy, 1.2 for galvanized), concrete density \(k_d\) (1.2 for semi-low-density, 1.3 for low-density), bar size \(k_s\) (0.8 for 20M and smaller) and reinforcement grade \(k_g\) (1.0 up to 400 MPa, \(f_{yg}/900 + 0.56\) above). They replace k1 to k4 from A23.3-19.
A horizontal bar with more than 300 mm of fresh concrete placed below it. Bleed water and air collect under such bars, so CSA A23.3-24 Table 12.2 gives them a bar location factor of 1.3.
Under CSA A23.3-24 Cl. 12.2.3, use 0.45 when the member has minimum stirrups or ties throughout the development length, or when the bars have at least 2 bar diameters of clear spacing and 1.5 bar diameters of clear cover. Everything else takes 0.6.

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Calculations referenced in this post

Development length in tension

Required Ld for bars in tension to CSA A23.3-24 Cl. 12.2

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