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[UPDATED] Design for Shear Using Bent-up Bars

Every reinforced concrete element subjected to flexure is always accompanied by a shear force. This is a force that results in diagonal tension in concrete, consequently leading to failure via the formation of cracks. This cracks would normally propagate from the supports and extend into the beam, ultimately leading to failure. To avoid a shear failure in a flexural reinforced concrete element, shear reinforcements are normally provided.

Fig. 1: Shear Failure in Concrete Beams

The conventional method of designing against shear in reinforced concrete structures is via the provision of shear-links (stirrups). These are steel bars vertically placed around the tensile reinforcement at suitable spacings along the length of the concrete element. However, in certain instances, shear cannot be resisted via the use of shear-links alone, especially where the magnitude of the applied shear-force is so enormous. In such cases, a very effective way of resisting shear in reinforced concrete is by combining the shear-links with the provision of bent-up bars.

Bent-up bars, although very effective are not very popular compared to shear-links, this is usually due to the increased cost of forming and fixing the reinforcement.

Basis of Bent-up Bars

Consider the simple case of a simply supported beam, subjected to a uniformly distributed loading. As the bending moment decreases to zero, the shear force increases towards the support. It, therefore, follows that the flexural tensile reinforcement is useless towards the support, hence it is normal to curtail the tensile flexural reinforcement. However, instead of curtailing the tension reinforcement towards the supports, they can be bent-up (Figure 2) to cross a potential shear crack, and thus assist in resisting the shear force.

bent up bars for simply supported
Fig. 2: Bent-up bars resisting shear force (simply supported end).

The same ideology can be extended to concrete elements with continuous support as well. However, in the case of continuous supports, both bending moment and shear force increase towards the supports. Hence with continuous supports, the tension reinforcement is bent-down to act as shear reinforcement as the bending moment decreases away from the supports (Figure 3).

describes bent-up bars in a continuous support
Fig. 3: Bent-down bars resisting shear force (continuous support).

This is the basis for using bent-up or bent-down bars. However, quite often there might not be sufficient bars to bend in order to comply with code specifications on minimum spacings.

Bent-up Bars Code Equation

The equations presented in this post only covers the use of bent-up bars only as shear reinforcement within concrete structures. It doesn’t extend to the use of shear links which is the most common method of resisting shear. Although the design equations presented are quite similar to those used for shear links, the design for shear using links have been deliberately ignored as most readers are quite familiar with the procedure.

Figure 4 shows the idealized model for the design in the Eurocode, a composite truss consisting of bent up bars and concrete struts. Assuming z is the lever arm, i.e. the distance between the compression and longitudinal reinforcement acting as tension chord. The concrete struts and the bent up bars are inclined at angles θ and α respectively to the horizontal.

Model for bent-up bars
Fig. 4: Composite truss with bent-up bars and concrete struts.

Taking a section parallel to the struts as shown in Figure. 5 the number of bent-up bars is z(cot θ + cot α)/s, where s is the spacing of the bent-up bars.

Compressive strut parallel for bent-up bars
Fig. 5: A section parallel to the concrete struts.

The vertical component of the bars is, therefore, the shear resistance VRd,s of the bent-up given as:

{ V }_{ Rd,s }=\frac { z\left( cot\theta +cot\alpha  \right)  }{ s } { A }_{ sw }{ f }_{ ywd }sin\alpha

Where, Asw is the total area of bent-up bars or  and fywd is the design yield stress of the bent-up bars

Similarly, taking a section perpendicular to the struts, the maximum shear strength in a concrete section with bent-up bars can be derived and shown as:

{ V }_{ Rd,max }={ \alpha  }_{ cw }{ b }_{ w }z{ v }_{ 1 }{ f }_{ cd }\frac { \left( cot\theta +cot\alpha  \right)  }{ \left( 1+{ cot }^{ 2 }\theta  \right)  }
{ \alpha  }_{ cw }=1.0;\quad { v }_{ 1 }=0.6;
z=0.9d;\quad { f }_{ cd }=\frac { { f }_{ ck } }{ { \gamma  }_{ c } }

Design Procedure

  • Verify that VRd, max > VEd
  • Design shear reinforcement
  • Verify serviceability & detailing requirement

Worked Example

A concrete beam of size (225 x 600) is required to resist a design shear force of VEd =  565kN using a combination of shear links and system of bent-up bars. The shear links are to be Y10-200 mm centres, while the bent up bars are to be Y16 bars bent in pairs at an angle of 45 and at a spacing of 750mm centres. Verify the adequacy of the system of bent-up bars in resisting 50% of the design shear force. Assuming a concrete of class C25/30 and steel bars of grade 460Mpa.

Step 1: Verify that VRd,max > VEd

{ V }_{ Ed }=\left( 0.5\times 565 \right) =282.5kN 
Assuming\quad cot\theta =2.5
cot\alpha =1\quad i.e\quad \alpha =45^{ \circ  }
{ V }_{ Rd,max }={ \alpha  }_{ cw }{ b }_{ w }z{ v }_{ 1 }{ f }_{ cd }\frac { \left( cot\theta +cot\alpha  \right)  }{ \left( 1+{ cot }^{ 2 }\theta  \right)  }
=225 (0.9\times 600) 0.6 \frac { 25 }{ 1.5 }  \frac { \left( 2.5+1 \right)  }{ \left( 1+{ 2.5 }^{ 2 } \right)  } \\=586.6kN
 { V }_{ Rd,max }>{ V }_{ Ed }\quad \\=\left( 586.6kN>282.5kN \right) \quad O.k

Step 2: Design Shear Reinforcement

{ A }_{ sw }=2\times 201=402{ mm }^{ 2 }
cot\theta =2.5;\quad \alpha =45^{ \circ  }
 z=0.9d=\left( 0.9\times 600 \right) =540mm
{ f }_{ ywd }=460Mpa;\quad s =600mm
{ V }_{ Rd,s }=\frac { z\left( cot\theta +cot\alpha  \right)  }{ s } { A }_{ sw }{ f }_{ ywd }sin\alpha
=\frac { 540\left( 2.5+1 \right)  }{ 750 }  402\times 460\times 0.7071\times { 10 }^{ -3 }
=329.5kN>282.5kN\quad O.k

Step 3: Serviceability Requirement

 s\le 0.75d\left( 1+cot\alpha  \right) \\=0.75\times 600\times \left( 1+1 \right)
=900mm>750mm\quad o.k

Also see: Design for Punching Shear in Concrete Elements

Thank You!!!

Published by Omotoriogun Victor

A dedicated, passion-driven and highly skilled engineer with extensive knowledge in research, construction and structural design of civil engineering structures to several codes of practices

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