Shear Reinforcement
Shear reinforcement is provided in the form of stirrups to resist the shear and arrest the propagation of shear cracks.
In the beam design philosophy the flexure and shear are treated separately even though they act together on the beam. Hence the only function of shear reinforcement is to act against the shear action and these reinforcements are not considered to play any role in the flexure of the beam.
Modes of shear failure
Shear Flexural failure
The tensile stress trajectories shown in dotted line are nearly horizontal at the midspan as shown in figure
Hence the flexural cracks perpendicular to these trajectories appear and these are generally stopped by the longitudinal tension reinforcement provided.
Shear diagonal tension failure
Near the supports the compressive stress trajectories shown in continuous lines are predominant.
The nature of cracks in this region is inclined at almost 45 degrees hence they are called as diagonal tension cracks. To arrest these cracks a vertical reinforcement is necessary, which is provided in terms of stirrups.
Shear compression failure
Shear compression failure occurs due to localized concentration of shear stress in concrete beam particularly near the supports. In this mode the concrete fails by crushing mode of failure and then the cracks start to propagate through the body of the concrete.
Functions of Stirrups
The adjacent figure shows the propagation of shear cracks for a positive bending case where tension occurs at bottom and compression occurs at top face of the beam. These cracks propagate through the body of the beam and needs to be arrested by reinforcement. In this case stirrups provide the action to arrest the propagation of shear cracks.
Stirrups also have a confining effect in case of columns when they are provided according to ductile detailing code IS 13920:2016. This create a strong core which are helpful in case of seismic events.
Shear strength of RC Beam
The shear strength of beam depends on the percentage of tension reinforcement, width, effective depth, steel grade of stirrups and grade of concrete.
As the percentage of steel increases the design shear strength of the beam increases, since percentage of steel is used to calculate the design shear strength from Table 19 of IS 456:2000.
Maximum Spacing between Stirrups
Maximum spacing between stirrups is given in clause 26.5.1.5 of IS 456:2000. According to this clause the maximum spacing is the least of following three criterias:
- \frac{0.87 * f_{y} * A_{s v} * d}{V_{u s}}
- 0.75 * effective depth
- 300mm
Minimum shear reinforcement
Minimum shear reinforcement is provided when the concrete and the longitudinal reinforcement of the beam are sufficient and capable of providing shear resistance. In this case the minimum shear reinforcement is calculated as follows:
where Asv-min is the minimum shear reinforcement required.
Vumin = 0.4 b*d
sv = spacing of the stirrups
Type of stirrups in shear reinforcement
- Vertical stirrups are generally the most common practice to provide in beams. The shear force resisted by vertical stirrups is given as V_{us}=\frac{0.87 * f_{y} * A_{s v} * d}{s_{v}}.
- Inclined stirrups are when stirrups are inclined at an angle or different bars are are bent up at different cross sections along the length of the beam. Shear resisted by inclined stirrups is greater than shear resisted by vertical stirrups. The shear force resisted by inclined stirrups is given as V_{us}=\frac{0.87 * f_{y} * A_{s v} * d}{s_{v}}*(sin{\alpha}+cos{\alpha})
- Bent up bars are basically single or group of bars bent up at the ends near the supports. The shear resisted by bent up bars is given as V_{us}={0.87 * f_{y} * A_{s v} }*sin{\alpha}
Procedure for the shear design of RC beam
Step 1: Find Nominal Shear Stress \tau_{v} ( Actual shear stress coming on member)
\tau_{v}=\frac{V_{u}}{b d}<\tau_{c-\max }Step 2: Find Shear stress taken by concrete \tau_{c} if No Shear Reinforcements are provided
This is dependent on percentage of tension reinforcement and grade of concrete and found using table 19 IS 456:2000.
Step 3: If \tau_{v}<\tau_{c} minimum shear reinforcement is provided as described in above section.
If \tau_{v}>\tau_{c} then
V_{us}=\frac{0.87 * f_{y} * A_{s v} * d}{s_{v}}where V_{us}=V_{u}-\tau_{c}*b*d
Step 4: Spacing is given as mentioned in section 6 of this article.
