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Development Length and Lap Splices in NSCP 2015, Step by Step

By 7 min readUpdated

"Forty times the bar diameter" is the lap length many of us learned on site. It is easy to remember, and sometimes it is close. But NSCP 2015 gives a proper way to compute it, and the answer can be much shorter or much longer than 40db depending on the concrete, the cover, the bar position and the ties.

This article explains what development length is, then works through one 20 mm bar with every step shown.

What development length means

A bar can only reach its yield stress fy if it is gripped by enough concrete. The grip comes from the ribs of the bar bearing on the concrete around it. Development length ld is the embedded length needed so the bar yields before it pulls out or splits the concrete.

Wherever a bar must carry its full stress, for example at the face of a support or at the point of maximum moment, it needs at least ld of embedment beyond that point, straight, hooked or headed.

A lap splice is two bars overlapping so that one passes its force to the other through the concrete. Its length is based on ld.

The example

  • Bar: 20 mm, in tension, bottom bar (not a "top bar")
  • f'c = 28 MPa, fy = 420 MPa, normal-weight concrete, uncoated bars
  • Clear cover 40 mm, clear spacing between bars 75 mm
  • Ties: 10 mm, 2 legs, @ 150 mm, 3 bars being developed across the tie

Common numbers: √28 = 5.29 MPa (under the 8.3 MPa cap of 425.4.1.4).

Step 1. The modification factors (Table 425.4.2.4)

Factor What it accounts for Value here
ψttop bar (more than 300 mm of fresh concrete below it)1.0 (bottom bar)
ψeepoxy coating1.0 (uncoated)
ψsbar size: 20 mm and smaller get a reduction0.8
λlightweight concrete1.0 (normal-weight)

Step 2. Confinement term

cb is the smaller of the cover to the bar centre and half the centre-to-centre spacing:

cb = min(40 + 20/2, (75 + 20)/2) = min(50, 47.5) = 47.5 mm

The ties add confinement through Ktr (Eq. 425.4.2.3b):

Atr = 2 × 78.5 = 157.1 mm²
Ktr = 40 × Atr / (s × n) = 40 × 157.1 / (150 × 3) = 13.96 mm

(cb + Ktr) / db = (47.5 + 13.96) / 20 = 3.07 → capped at 2.5

Cross-section for cb and Ktr, 20 mm bar example To-scale partial beam cross-section with three 20 mm bottom bars at 40 mm clear cover and 75 mm clear spacing, inside a 10 mm two-leg tie at 150 mm. cb is the smaller of 40 + 20/2 = 50 mm (cover to bar centre) and (75 + 20)/2 = 47.5 mm (half the centre-to-centre spacing), so cb = 47.5 mm. Atr = 2 x 78.5 = 157.1 mm2, Ktr = 40 x 157.1 / (150 x 3) = 13.96 mm, and (cb + Ktr)/db = (47.5 + 13.96)/20 = 3.07, capped at 2.5. CROSS-SECTION FOR cb AND Ktr to scale, mm (b) 47.5 (a) 50 3 bars Ø20, n = 3 splitting plane tie Ø10, 2 legs @ 150 40 40 20 75 20 75 20 40 Clear cover 40 and clear spacing 75 are measured to the developed bars. Both tie legs cross the splitting plane, so Atr counts 2 legs. cb = the smaller of (a) cover to bar centre 40 + 20/2 = 50 mm (b) half the c/c spacing (75 + 20)/2 = 47.5 mm cb = 47.5 mm (b) governs Ktr from the ties Ø10, 2 legs @ 150 mm, n = 3 Atr = 2 × 78.5 = 157.1 mm² Ktr = 40 Atr / (s n) = 40 × 157.1 / (150 × 3) = 13.96 mm (cb + Ktr)/db = (47.5 + 13.96)/20 = 3.07 CAPPED AT 2.5
Figure 1. Cross-section of the example, to scale: three 20 mm bottom bars at 40 mm clear cover and 75 mm clear spacing inside 10 mm two-leg ties at 150 mm. cb is the smaller of 50 mm and 47.5 mm, Ktr = 13.96 mm, and (cb + Ktr)/db = 3.07 is capped at 2.5.

Step 3. Development length, detailed equation (Eq. 425.4.2.3a)

ld = [ fy / (1.1 λ √f'c) × (ψt ψe ψs) / ((cb + Ktr)/db) ] × db
ld = [ 420 / (1.1 × 1.0 × 5.29) × (1.0 × 1.0 × 0.8) / 2.5 ] × 20
ld = 72.16 × 0.32 × 20 = 461.8 mm (≥ 300 mm minimum, OK)

That is about 23db, much less than 40db, because the cover and ties are good and the bar is small.

Step 4. The simple table (Table 425.4.2.2), for comparison

If the clear spacing is at least db, the cover is at least db, and minimum ties are provided, NSCP allows a quick formula without working out cb and Ktr. For 20 mm and smaller bars:

ld = fy ψt ψe / (2.1 λ √f'c) × db = 420 / (2.1 × 5.29) × 20 = 755.9 mm

That is 38db, close to the old "40db" rule. The table is quick and safe, but it costs you over 60% more length than the detailed equation here. On a big job, that is a lot of steel.

Step 5. Tension lap splices (425.5.2)

NSCP has two classes:

  • Class A = 1.0 ld, allowed only if As provided ≥ 2 × As required over the whole lap and no more than half the bars are spliced in the lap length.
  • Class B = 1.3 ld, everything else. This is the usual case.

With ld = 461.8 mm:

Class A = 461.8 mm; Class B = 1.3 × 461.8 = 600.4 mm (both ≥ 300 mm)

Use Class B, 600 mm (30db), unless you have checked the Class A conditions.

Tension lap splice, Class A and Class B, 20 mm bars Two pairs of 20 mm bars in contact, drawn with their lap lengths to scale. Class A = 1.0 ld = 461.8 mm, allowed only if As provided is at least 2 x As required over the whole lap and no more than half of the bars are spliced in the lap length. Class B = 1.3 ld = 1.3 x 461.8 = 600.4 mm, the usual case: use 600 mm, which is 30 bar diameters. Both laps exceed the 300 mm minimum. TENSION LAP SPLICE: CLASS A AND CLASS B Ø20 bars, ld = 461.8 mm Class A = 1.0 ld ONLY IF BOTH CONDITIONS MET Allowed only if As provided ≥ 2 × As required over the whole lap and no more than half of the bars are spliced in the lap length. Class A lap = 461.8 mm bar 1, Ø20 bar 2, Ø20 Class B = 1.3 ld USUAL CASE: USE 600 mm Everything else. This is the usual case. Class B lap = 1.3 × 461.8 = 600.4 mm, use 600 mm (30db) bar 1, Ø20 bar 2, Ø20 300 mm minimum Lengths along the bars and bar diameters to scale. f'c 28 MPa, fy 420 MPa; ld from Eq. 425.4.2.3a.
Figure 2. Class A and Class B tension laps for the 20 mm bar, lengths to scale: Class A = 1.0 ld = 461.8 mm only when both conditions are met; Class B = 1.3 ld = 600.4 mm, so use 600 mm.

When bars of different size are lapped, the lap is the larger of ld of the larger bar and the lap length of the smaller bar (425.5.2.2).

Step 6. Bars in compression (425.4.9)

ldc = max(0.24 fy / (λ √f'c) × db, 0.043 fy × db)
= max(0.24 × 420 / 5.29 × 20, 0.043 × 420 × 20) = max(381.0, 361.2) = 381 mm

Compression lap for fy ≤ 420 MPa (425.5.5.1): 0.071 × fy × db = 0.071 × 420 × 20 = 596.4 mm (≥ 300 mm). If f'c is below 21 MPa, the compression lap is increased by one-third.

Step 7. Standard hooks (425.4.3)

When there is no room for a straight bar, as at an exterior beam-column joint, use a standard hook:

ldh = 0.24 fy ψe ψc ψr / (λ √f'c) × db

Side cover here is at least 65 mm (and the tail cover of a 90-degree hook at least 50 mm), so ψc = 0.7. With no ties enclosing the hook, ψr = 1.0:

ldh = 0.24 × 420 × 0.7 / 5.29 × 20 = about 266 mm (≥ 8db = 160 mm and ≥ 150 mm, OK)

Hook geometry for a 20 mm bar: inside bend diameter 6db = 120 mm; tail 12db = 240 mm for a 90-degree hook, 4db (min 65 mm) = 80 mm for a 180-degree hook.

Standard 90-degree and 180-degree hooks, 20 mm bar Both standard hooks for the 20 mm bar drawn to the same scale. A beam stub enters each joint from the left. The development length ldh, about 266 mm, is measured from the critical section to the outside of the bend. Inside bend diameter 6db = 120 mm. The 90-degree hook has a 12db = 240 mm tail with at least 50 mm cover beyond its end; the 180-degree hook has a 4db = 80 mm tail (minimum 65 mm). ldh = 0.24 x 420 x 0.7 / 5.29 x 20 = about 266 mm, which is at least 8db = 160 mm and at least 150 mm: OK. STANDARD HOOKS, Ø20 BAR, BOTH TO THE SAME SCALE mm D = 6db = 120 ldh ≈ 266 12db = 240 ≥ 50 face 90-degree hook critical section D = 6db = 120 ldh ≈ 266 4db (min 65) = 80 180-degree hook critical section side cover ≥ 65 mm (normal to the plane of the hook) ldh = 0.24 fy ψe ψc ψr / (λ √f'c) × db = 0.24 × 420 × 0.7 / 5.29 × 20 ≈ 266 mm ψc = 0.7: side cover ≥ 65 mm and 90-degree tail cover ≥ 50 mm. ψr = 1.0: no ties enclosing the hook. 266 ≥ 8db = 160 mm and ≥ 150 mm OK
Figure 3. Standard 90- and 180-degree hooks for the 20 mm bar, both to the same scale: ldh of about 266 mm to the outside of the bend, inside bend diameter 6db = 120 mm, tail 12db = 240 mm or 4db = 80 mm.

How much do the factors matter?

Change only two things: make it a 32 mm epoxy-coated top bar (more than 300 mm of concrete cast below it).

  • ψt = 1.3, ψe = 1.5, but their product is capped at 1.7
  • ψs = 1.0 (25 mm and larger)
  • cb = 53.5 mm, (cb + Ktr)/db = 2.11

ld = 72.16 × 1.7 × 1.0 / 2.11 × 32 = 1,862 mm, about 58db

The same "40db" rule would be short by almost 600 mm. This is why a blanket rule is risky: it is too long for small bottom bars and too short for large top bars.

Computed lengths in bar diameters against the 40db site rule Horizontal bar chart in multiples of the bar diameter db. 20 mm bottom bar, ld by the detailed equation: 461.8 mm, about 23db. 20 mm Class B tension lap: 600 mm, 30db. 20 mm compression lap: 596.4 mm, about 30db. 20 mm bottom bar, ld by the simple table: 755.9 mm, 38db. 32 mm epoxy-coated top bar, ld: 1,862 mm, about 58db. A dashed line marks the 40db site rule: too long for the small bottom bar and too short for the large epoxy-coated top bar. LENGTH IN BAR DIAMETERS AGAINST THE 40db SITE RULE 0 10db 20db 30db 40db 50db 60db length in bar diameters (db) 20 mm bottom bar ld, detailed equation 461.8 mm · 23db 20 mm bar Class B tension lap 600 mm · 30db 20 mm bar compression lap 596.4 mm · 30db 20 mm bottom bar ld, simple table 755.9 mm · 38db 32 mm epoxy-coated top bar ld, detailed equation 1,862 mm · 58db 40db site rule 40db TOO LONG for the 20 mm bottom bar with good cover and ties (23db needed) 40db TOO SHORT for the 32 mm epoxy-coated top bar (58db needed)
Figure 4. The example lengths in bar diameters against the 40db site rule: 23db to 38db for the 20 mm bar, 58db for the 32 mm epoxy-coated top bar.

Quick site guide

Situation Rough length (f'c 28, fy 420, good cover and ties)
20 mm bottom bar, ldabout 460 mm (23db)
20 mm, Class B tension lapabout 600 mm (30db)
20 mm, compression lapabout 600 mm (30db)
20 mm, 90-degree hookabout 270 mm embedment + 240 mm tail
32 mm epoxy top bar, ldabout 1,860 mm (58db)

These are for this example only. Lower f'c, smaller cover or closer bar spacing all increase the length. Always compute for your own case.

The section-details page of the RHCES Development Length Calculator for this same 20 mm example: ld 462 mm, Class B lap 600 mm, and both standard hooks with ldh 266 mm.
Figure 5. The section-details page of the RHCES Development Length Calculator for this same 20 mm example: ld 462 mm, Class B lap 600 mm, and both standard hooks with ldh 266 mm.

General information only. Follow the approved structural plans and the engineer of record.