A corbel is a short bracket sticking out of a column. It is too short to bend like a beam: the load runs straight down to the column in a sloping concrete strut, held at the top by a horizontal steel tie. This strut-and-tie model gives the tie steel; the shear, the links, the bearing under the load and the anchorage of the tie finish the design. Every number below comes from the free STATIX corbel calculator with the inputs in step 1.
1. The problem
Input
Value
Design code
SANS 10100-1
Element nib: cl.6.2.4.3
Single corbel
Width b
400 mm
Depth at the face / front
450 mm
Depth at the outer edge mm
250 mm
Projection
300 mm
Load position av, load line to the face
150 mm
Cover main bars
40 mm
Vertical Fv, ULS
300 kN
Bearing type
Bedded (0.6 fcu)
Pad along / across
150 mm
Pad width across the corbel mm
300 mm
Concrete fcu
40 MPa
Steel fy
450 MPa
Main tie number / dia
4 mm
Main bar diameter mm
16 mm
Horizontal links number / dia / legs
3
Link diameter mm
10
Legs per link
2
Exposure
Moderate
Fire rating
1 h
2. Is it a corbel?
The rules apply when the load sits closer to the column face than the effective depth (a_v < d), and the outer end is at least half as deep as the face.
a_v = 150 mm, d = 402 mm, a_v / d = 0.373
Short cantilever: av less than d at the face:
[SANS 10100-1 cl.6.2.4.2.1 (p.143)]
av < d (otherwise design as a cantilever beam, cl.4.3)
150 ≤ 402 mm OK
Depth at the outer end >= 1/2 depth at the face:
[SANS 10100-1 cl.6.2.4.2.1 (p.143)]
hOuter >= h/2
225 ≤ 250 mm OK
3. The strut-and-tie model
The compression in the concrete runs from the load down to the bottom of the column face. Its depth x sets the lever arm z and the slope β of the strut. Equilibrium at the load point then gives the strut force and the tie force; a horizontal force on the corbel adds straight to the tie.
x = 126.6 mm (x/d = 0.315), z = 345.0 mm
β = 66.5°
strut = 327.1 kN, tie = 130.4 kN (F_v 130.4 + F_h 0.0)
steel stress f_s = 391.5 MPa
4. Tie steel and the strut
The tie steel must carry the tie force at the steel stress the strains allow, and lie between a minimum and a maximum. The concrete strut must carry the vertical load.
Concrete strut (stress block 0,45 fcu over 0,9x, x <= d/2):
[SANS 10100-1 cl.6.2.4.2.3 (p.143), 4.3.3.1 b) (p.22)]
Fv <= strut capacity at x = d/2 (or its peak below d/2)
300 ≤ 509 kN OK
Main tie steel (carrying Fv and Fh), fs = 391.5 MPa:
[SANS 10100-1 cl.6.2.4.2.3 (p.143), 4.3.3.1 d) fig.2]
As >= Ft/fs
333 ≤ 804 mm² OK
Main tie >= 0,4 % of the section at the face (b h):
[SANS 10100-1 cl.6.2.4.2.3 (p.143)]
As >= 0,004 b h
720 ≤ 804 mm² OK
Main tie <= 4 % of the gross section:
[SANS 10100-1 cl.4.11.5.1 (p.96)]
As <= 0,04 b h
804 ≤ 7,200 mm² OK
5. Shear
Close to a support the concrete takes much more shear than in a long span, because the load goes straight into the support by arching: the shear strength is enhanced by 2d / a_v. An upper limit applies at the face.
v_c = 0.497 MPa → enhanced 2.663 MPa
Shear stress at the face:
[SANS 10100-1 cl.6.2.4.2.2 (p.143), 4.3.4.2.1 (p.29)]
v < lesser of 0,7 sqrt(fcu) and 4 MPa
1.87 ≤ 4.00 MPa OK
Shear: enhanced concrete stress vc 2d/av (100As/bd = 0.5):
[SANS 10100-1 cl.4.3.4.2.1 (p.29), eq.(2) cl.4.3.4.1 (p.27)]
v <= vc 2d/av: no shear reinforcement needed by 4.3.4.2.2
1.87 ≤ 2.66 MPa OK
6. Horizontal links
Closed horizontal links in the upper two-thirds of the depth control the splitting of the strut. They must give at least half the area of the main tie, stay inside the sloping soffit and leave room for the concrete.
Horizontal links >= 1/2 of the main tie (3 x 2 legs Y10):
[SANS 10100-1 cl.6.2.4.2.5 (p.144)]
sum of link legs crossing the face >= 0,5 As,main (provided)
402.1 ≤ 471.2 mm² OK
Horizontal links stay inside the sloping soffit and cross the strut (worst link at y 316):
[SANS 10100-1 cl.6.2.4.2.5 (p.144), 4.11.2 cover]
link end x (cover + dia/2 inside the soffit) > strut line x at the link level
36.4 ≤ 119.9 mm OK
Clear gap between horizontal links (spacing 89.333 mm over the upper 2d/3):
[SANS 10100-1 cl.4.11.8.1.2.1 (p.106)]
clear gap >= hagg + 5
25.0 ≤ 79.3 mm OK
7. Bearing under the load
The load comes onto the corbel through a bearing pad. The pressure under the pad is limited, the pad must be wide enough, and it must sit far enough back from the front edge so the corner does not spall off.
Bearing stress under the pad (bedded, effective length 250 mm):
[SANS 10100-1 cl.6.2.4.4.3, 6.2.4.4.4 (p.145)]
bedded: 0.6 fcu (weaker surface); Leff = least of L, L/2 + 100, 600
8.0 ≤ 24.0 MPa OK
Net bearing width (pad along the projection), isolated unit +20 mm:
[SANS 10100-1 cl.6.2.4.4.2 (p.144)]
>= max(40, Fv/(Leff fb)) + 20
70.0 ≤ 150.0 mm OK
Pad clear of the spalling zone at the front edge (edge distance 75 mm):
[SANS 10100-1 cl.6.2.4.5.1 (p.145)]
e) RC less than 300 mm deep at the outer edge: nominal cover
40.0 ≤ 75.0 mm OK
8. Anchoring the tie
The tie force must be developed right at the front of the corbel, where there is no room for a bar to anchor by bond. So the main bars are welded to a transverse bar (or looped). At the column end they must anchor into the column.
Welded transverse bar of equal strength (same grade):
[SANS 10100-1 cl.6.2.4.2.3 (p.143)]
transverse bar dia >= main bar dia
16 ≤ 20 mm OK
Weld of the main bars to the transverse bar: proved weld, joint strength 0,8 fy/1,15:
[SANS 10100-1 cl.6.2.4.2.3 (p.143), 4.11.6.7.1-4.11.6.7.3 (p.99)]
tie force <= n As 0,8 fy / 1,15 (all tie bars welded at one section: the 100 % option of 4.11.6.7.3 does not apply; no cold-worked bars, 4.11.6.7.1)
130 ≤ 252 kN OK
Main bar anchorage into the column from its depth 600 mm (90 deg bend down at the far face (cover 40, r 48): straight 496 + bend 192, leg 4 dia; fbu 2.38, top bar -30 %):
[SANS 10100-1 cl.4.11.6.3, table 24 (p.97), 4.11.6.8.2 b) (p.100)]
available anchorage >= 0,87 fy dia/(4 fbu)
658 ≤ 688 mm OK
Horizontal links fully anchored: 135 deg round a bar of dia 16 (>= link dia 10), extension 100 mm (>= 8 dia, 135 deg taken by the 90 deg rule):
[SANS 10100-1 cl.6.2.4.2.5 (p.144), 4.11.6.4 (p.97-98)]
extension >= 80 mm and corner bar dia >= link dia
80 ≤ 100 mm OK
9. Cover, fire and fit
The bars must fit across the width, and the cover must suit the exposure and the fire rating.
Main bars fit across b: clear gap:
[SANS 10100-1 cl.4.11.8.1.2.1 (p.106)]
clear gap >= hagg + 5
25 ≤ 85 mm OK
Nominal cover >= bar diameter and >= max aggregate:
[SANS 10100-1 cl.4.11.2.4, 4.11.2.5 (p.92)]
>= max(bar dia, hagg)
20 ≤ 40 mm OK
Durability cover: moderate exposure, concrete class 40 (fcu 40), least of top and side cover:
[SANS 10100-1 cl.4.11.2.2, 4.11.2.10 (p.92) -> SABS 0100-2 (not on disk); cover table: SANS 2001-CC1:2007 cl.4.4.3.1 table 3 (p.23)]
>= 30 mm to the outside of any bar or link (Moderate: surfaces protected by the superstructure, enclosed surfaces, not moistened by condensation)
30 ≤ 40 mm OK
Fire resistance 1 h: average cover to the main tie (nearest exposed face):
[SANS 10100-1 cl.6.2.4.2.1 (p.143), 7.2.1-7.2.4 (p.164), table 43 a) siliceous aggregate (p.165)]
>= 25 mm (siliceous aggregate, the row for normal-density concrete; no plaster counted)
25 ≤ 40 mm OK
Fire resistance 1 h: corbel width:
[SANS 10100-1 cl.6.2.4.2.1 (p.143), 7.2.1-7.2.4 (p.164), table 43 a) siliceous aggregate (p.165)]
b >= 110 mm
110 ≤ 400 mm OK
A nib (a continuous ledge) follows the same model per metre run; the calc sheet shows each formula with its clause.
Run this example yourselfOpen the free STATIX corbel calculator and press Load the example (300 kN at 150 mm) under "How it is calculated". Change any value and every step updates, with a calc sheet. Open the corbel calculator
A worked example for learning, not a design. Check every value against the current code and your own project, and have the design checked and signed by a registered professional engineer.
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