Concrete Shear Design |

Civil's Guide

Shear Design

Concrete Shear

When concrete is subjected loading from it’s self-weight, permanent and variable loads, diagonal cracks forms in beams and additional reinforcement is required to minimise cracking. In slabs and foundations, large loading can result in high localised “punching shear” occuring around supports(columns), which may also require additional reinforcement in those regions. See the image below which shows the types of cracks forming in a typical beam.

Shear is carried out at ULS only (i.e, SLS checks not required).
Shear failure is much more complex than flexural failure.
It is still a subject of research.
However, current theories are adequate for design.

Concrete Sections that do no require Shear reinforcement

Lightly loaded floor slabs, pad foundations etc
Where shear forces are small the concrete section on its own may have sufficient shear capacity (VRd,c) to resist the ultimate shear force (Ved)
In beams a minimum amount of shear reinforcement will usually be provided.
The shear capacity of the concrete, VRd,c in such situations is given by an empirical expression: \(V_{RD,c} = (0.12k(100\rho f_{ck})^{1/3})b_{w}d\)
Where \(k = (1 + \sqrt{\frac{200}{d}})\leq2.0\)
and \(\rho = \frac{A_{sl}}{b_wd} \leq 2.0 \)
with a minimum value of: \(V_{Rd,c} = (0.035k^{3/2}f_{ck}^{1/2}) b_{w}d \)

Variable Strut Inclination method

When designing for shear in beams, the applied loading/actions are represented by an analogous truss. The concrete acts as the top compression member and at an angle \(\theta\). The bottom chord is the tension steel (main steel bars) and the steel links (stirrups) will act as tension members. In subsequent calculations when determining the shear capacity using shear links, the concrete does not contribute to the shear capacity of the beam (i.e, capacity is determined from steel links).

Analysis of the beam to find the ULS Shear resistance is carried out in three stages:

1. Check the compressive strength of the diagonal concrete strut and its angle \(\theta\)
2. Calculate required shear reinforcement \(A_{sw}/s\)
3. Calculate additional area of tension steel \(A_{sl}\) required in the bottom chord.

1. Diagonal compressive strut

Excessive compressive stresses must not occur in the diagonal strut:

The effective cross sectional area of concrete acting as the diagonal strut is taken as: \(b_w \times z\cos\theta\)
and the design concrete stress: \(f_{cd} = f_{ck}/1.5\)
The ultimate strength of the strut = ultimate design stress × area = \((f_{ck}/1.5) \times (b_w \times z\cos\theta)\)
and its vertical component = \([(f_{ck}/1.5) \times (b_w \times z\cos\theta)] \times \sin\theta\)
so that \(V_{Rd,max} = (f_{ck} b_{w} z \cos\theta sin\theta)/1.5 \)
By conversion of the trigometric functions this can be expressed as: \(V_{Rd,max} = \frac{f_{ck}b_wz}{1.5(\cot\theta + \tan\theta)}\)
In EC2 this equation is modified b ythe inclusion of a strength reduction factor \(v_1\) for concrete cracked in shear
Thus \(V_{Rd,max} = \frac{f_{ck}b_wzv_1}{1.5(\cot\theta + \tan\theta)}\)
Where the strength reduction factor takes the value \(v_1=0.6(1-f_{ck}/250)\)
and z = 0.9d then the above becomes:
\(V_{Rd,max} = \frac{0.9dxb_w\times0.6(1-f_{ck}/250)f_{ck}}{1.5(\cot\theta + \tan \theta)}\)
\(=(0.36b_wd(1-f{ck}/250)f_{ck})/(\cot\theta+\tan\theta)\) …. Eqn 1

EC2 Limits

EC2 limits \theta to a value between \(22^{\circ}\) and \(45^\circ\)
(i) With \(\theta = 22^\circ\) (this is the usual case for udl loads) from Eqn 1: \(V_{Rd,max(22)} = 0.124b_wd(1-f_{ck}/250)f{ck}\)
If \(V_{Rd,max.(22)}\) \(<\) \(V_{Ed}\) then a larger \(\theta\) must be used so that the diagonal concrete strut has a larger vertical component to balance \(V_{Ed}\).
(ii) With \(\theta\) = \(45^\circ\) (the maximum value of \(\theta\) as allowed by EC2) again from Eqn 1:
\(V_{Rd,max(45)} = 0.18b_wd(1-f_{ck}/250)f_{ck}\)
Which is the upper limit on the compressive strength of the concrete diagonal member in the analogous truss.
If \(V_{Rd,max.(45)} < V_{Ed}\) then a larger concrete section is required
(iii) With  = between 22 and 45 The required value of  can be obtained by equating VEd to VRd,max and solving  in Eqn 1 as follows:
\(V_{ED}=V_{Rd,Max}=\frac{0.36b_wd(1-f_{ck}/250)f_{ck}}{\cot\theta + \tan\theta}\)
and \(1/(\cot\theta + tan\theta)=\sin\theta \times \cos\theta = 0.5\sin2\theta\)
Therefore by substitution \(\theta = 0.5\sin^{-1}[V_{ED}/0.18b_wd(1-f_{ck}/250)f{ck}]\leq 45^\circ\)
This can also be shown as \(\theta=0.5sin^{-1}[V_{Ef}/V_{Rd,max.(45)}]\leq 45^\circ\)
This calculated angle \(\theta\) can now be used to determine \(\cot\theta\) when calculating the required shear reinforcement.

2. Vertical Shear Reinforcement

Shear is resisted by shear links with no contribution from the concrete.
Using the method of sections (cut at X-X), the force in the vertical link member (\V_{wd}\) must equal the shear force \(V_{ED}\), that is
\(V_{wd} = V_{Ed} = 0.87f_{yk}A_{sw}\)
If the links are spaced at a distance s apart, then the force in each link is reduced proportionately and is given by
\(V_{wd} = B_{Ed}=0.87\frac{A_{sw}}{s}zf_{yk}\cot\theta = 0.87\frac{A_{sw}}{s}0.9df_{yk}\cot\theta\)
thus rearranging: \(\frac{A_{sw}}{s}=\frac{V_{Ed}}{0.78df_{yk}\cot\theta}\)
EC2 specifies a minimum area of links of: \(\frac{A_{sw,min}}{s}=\frac{0.08f_{ck}^{0.5}b_w}{f_{yk}}\)

3. Additional Longitudinal Force

Allowance must be made for the additional longitudinal force in the tension steel
Resolving forces horizontally (cut at Y-Y), the longitudinal component of the force in the compressive strut is given by :
Longitudinal Force = \((V_{Ed}/\sin\theta) \times \cos \theta\) = \(V_{Ed}\cot\theta\)
It is assumed that half this force is carried by the reinforcement in the tension zone of the beam, then: \(F_{td}=0.5V_{Ed}\cot\theta\)
This can be provided by additional longitudinal reinforcement above that required for bending reinforcement

Shear reinforcement - variable strut inclination method

Punching Shear

When looking at shear stresses in slabs, the loads will be distruted through the slab (general actions are uniformly distributed loads/pressure loads), which will have a ultimate shear force \(V_{Ed}\) less than the shear resistance of the concrete \(V_{Rd,c}\). Shear reinforcement is not necessarily required (check point loads as a seperate case). However, localised ‘punching shear’ forms around the column due to the concentrated shear loads at this area (reaction force from applied loading); these cases are critical and a basic control perimeter is 2.0d from the loaded area and will generally require shear reinforcement.

The equation for punching shear stress can be expressed as: \(\frac{\beta V_{Ed}}{U_id}\)

Where
\(\beta\) = 1.0 for columns with no eccentricity/continuity
\(\beta\) = 1.15 for internal columns
\(\beta\) = 1.4 for edge columns
\(U_i\) = 1.5 for corner columns
d = effective depth

When the applied shear stress is greater than the shear resistance of the concrete, shear reinforcement will be required. In general, the additional reinforcement is procured from manufactor reinforcement systems for this issue and they will undertake the calculations through their softwares.