A corbel is a very short structural cantilever member projecting from a wall or a column for the purpose of carrying loads. In reinforced concrete structures, corbels are cast monolithically with the walls or columns supporting them. They are found mainly in bridges, industrial buildings, and commercial buildings with precast construction.

If a structure is subjected to in-plane forces, then the stress distribution will consist of normal stresses in the two planes (Ïƒ_{x} and Ïƒ_{y}) and an accompanying shear stress *Ï„*_{xy}. These stresses will lead to two principal stresses Ïƒ_{1} and Ïƒ_{2}. If for instance Ïƒ_{1} is tensile and Ïƒ_{2} is compressive, then the external load on the structure can be idealized as being resisted by a combination of tensile and compressive stresses.

In reinforced concrete structures, the compressive stresses can be resisted by the concrete, while the tensile stresses can be resisted by the steel reinforcement. Generally, these are idealized by a series of concrete struts and steel ties, and forms the underlying principle behind the strut-and-tie method of design. Structures that can be designed using this method are pile caps, deep beams, corbels, half-joints, etc.

Under certain conditions, Eurocode 2 permits the use of strut and tie method for the analysis and design of corbels. According to clause 6.5.1 of EN 1992-1-1:2004, strut-and-tie method can be used where non-linear stress distribution exists (for example at supports, near concentrated loads or plain stress).

Strut-and-tie models utilize the lower-bound theorem of plasticity which can be summarized as follows: for a structure under a given system of external loads, if a stress distribution throughout the structure can be found such that;

(1) all conditions of equilibrium are satisfied and

(2) the yield condition is not violated anywhere,

Then the structure is safe under the given system of external loads.

When using the strutâˆ’and-tie method of design, because the given structure is replaced by a pin-jointed truss, different types of nodes occur where members meet. These nodal zones need to be carefully designed and detailed.

(a) **CCC Node**: If three compressive forces meet at a node, it is called a CCC node. According to Eurocode 2 equation (6.60), the compression stress in each strut is restricted to a maximum value of 1.0(1 â€“ f_{ck}/250)f_{cd}.

(b) **CCT node**: If two compressive forces and a tie force anchored in the node through bond meet, it is called a CCT node. According to Eurocode 2 equation (6.61), the compression stress in each strut is restricted to a maximum value of 0.85(1 â€“ f_{ck}/250) f_{cd}.

(c) **CTT node**: If two tensile forces at a compressive force meet at a node it is called a CTT node. According to Eurocode 2 equation (6.62),

the compression stress in each strut is restricted to a maximum value of 0.75(1 â€“ f_{ck}/250)f_{cd}.

The following conditions are to be applied in the design of corbels;

(1) Corbels with 0.4h_{c} < a_{c} < h_{c} may be designed using a simple strut and tie model.

(2) For deeper corbels (a_{c} < 0.4h_{c)}, other adequate strut and tie models may be considered.

(3) Corbels for which a_{c} > h_{c }may be designed as cantilever beams

(4) Unless special provision is made to limit horizontal forces on the support, or other justification is given, the corbel should be designed for the vertical force F_{v}, and a horizontal force H_{c} > 0.2F_{v} acting at the bearing area.

(5) The overall depth (h_{c}) of the corbel should be determined from considerations of shear.

(6) The local effects due to the assumed strut and tie system should be considered in the overall design of the supporting member.

(7) The detailing requirements shall be met.

**Design Example of a thick short corbel (a**_{c} < h_{c}/2)

_{c}< h

_{c}/2)

Design the corbel shown below. The bearing area of the support is 200 x 300 mm. The width of the corbel is 150 mm.

**Materials**

Concerete C30/35, f_{ck }= 30 N/mm^{2}; f_{yk} = 500 N/mm^{2}

Design strength of concrete f_{cd }= Î±_{cc}f_{ck}/Î³_{c} = (0.85 x 30)/1.5 = 17 N/mm^{2}

Design strength of steel f_{yd }= f_{yk}/Î³_{s} = 500/1.15 = 435 N/mm^{2}

**Compressive strength of nodes;**

**CCC nodes:**

Ïƒ_{1Rd,max} = 1.0 x (1 â€“ f_{ck}/250)f_{cd} = 1.0 x (1 â€“ 30/250) x 17 = 14.96 N/mm^{2}**CCT nodes:**

Ïƒ_{2Rd,max} = 0.85 x (1 â€“ f_{ck}/250)f_{cd} = 0.85 x (1 â€“ 30/250) x 17 = 12.716 N/mm^{2}**CTT nodes:**

Ïƒ_{3Rd,max} = 0.75 x (1 â€“ f_{ck}/250)f_{cd} = 0.75 x (1 â€“ 30/250) x 17 = 11.22 N/mm^{2}

**Actions**

F_{Ed} = 600 kN

Load eccentricity with respect to the column side: e = 125 mm

The beam vertical strut width is evaluated by setting the compressive stress equal to Ïƒ_{1Rd,max}:

*x _{1}* = F

_{Ed}/Ïƒ

_{1Rd,max}b = (600 x 10

^{3})/(14.96 x 350) = 114.6 mm

Node 1 is therefore located at

*x*/2 = 114.6/2 = 57.3 mm

_{1}Let the cover from the top of the cantilever to the reinforcements be 30 mm. Assuming 16 mm bars;

The effective depth d = 450 â€“ 35 â€“ (16/2) = 407 mm

The distance y_{1} of the node 1 from the lower border is evaluated setting the internal drive arm *z* equal to 0.8â‹…d (z = 0.8 x 407 = 325.6 mm):

y_{1} = 0.2d = 0.2 x 407 = 81.5 mm

Rotational equilibrium: F_{Ed}a = F_{c}z

a = ac + *x _{1}*/2 = 125 + 57.3 = 182.3 mm

600 x 10

^{3}x 182.3 = F

_{c}x 325.6

F

_{c}= F

_{t}= 335933.66 N = 335.93 kN

**Node 1 verification;**

Ïƒ = F_{c} / (bâˆ™2y_{1}) = (335.93 x 10^{3})/(350 x 2 x 81.5) = 5.89 N/mm^{2} < Ïƒ_{1Rd,max} (14.96 N/mm^{2})

**Main Top Reinforcement Design**

A_{s1} = F_{t}/f_{yd} = (335.93 x 10^{3})/435 = 772 mm^{2}

Provide 5H16 (A_{s,prov} = 1005 mm^{2})

**Shear reinforcement design**

The beam proposed in EC2 is indeterminate, then it is not possible to evaluate the stresses for each single bar by equilibrium equations only, but we need to know the stiffness of the two elementary beams shown below in order to make the partition of the diagonal stress [F_{diag} = F_{c}/cosÎ¸ = F_{Ed}/sinÎ¸] between them.

Based on the trend of main compressive stresses resulting from linear elastic analysis at finite elements, some researchers from Stuttgart have determined the two rates in which F_{diag} is divided, and they have provided the following expression of stress in the secondary reinforcement.

F_{wd} = [(2z/a â€“ 1)/(3 + F_{Ed}/F_{c})] x F_{c} = [(2(325.6/182.3) â€“ 1)/(3 + 600/335.93)] x 335.93 = 180.525 kN

A_{sw }= F_{wd}/f_{yd }= (180.525 x 10^{3})/435 = 415 mm^{2} > k_{1}A_{s,prov} = 0.25 x 1005 = 251.25 mm^{2}

Trying 2 legs of H10 mm, Provide 4 number of H10 mm stirrups (A_{sw,prov} = 628 mm^{2})

**Node 2 verification, below the load plate**

The node 2 is a tied-compressed node, where the main reinforcement is anchored; the compressive stress below the load plate is;

Ïƒ = F_{Ed} /(200 x 300) = (600 x 10^{3})/(200 x 300) = 10 N/mm^{2} < Ïƒ_{2Rd,max} (12.716 N/mm^{2}) Okay

**Detailing Requirements**

(1) The reinforcement, corresponding to the ties considered in the design, should be fully anchored beyond the node under the bearing plate by using U-hoops or anchorage devices, unless a length l_{b,net} is available between the node and the front of the corbel, l_{b,net} should be measured from the point where the compression stresses change their direction.

(2) In corbels with h_{c} > 300 mm, when the area of the primary horizontal tie A_{s} is such that (where A_{c }denotes the sectional area of the concrete in the corbel at the column), then closed stirrups, having a total area not less than |0.4|As, should be distributed over the effective depth d in order to cater for splitting stresses in the concrete strut. They can be placed either horizontally or inclined.