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Design guide

Shear design and links to Eurocode 2: a worked example

Shear and links to EN 1992-1-1, worked step by step: resistance without links, the variable strut inclination method, the crushing limit and minimum links.

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A beam reinforcement cage on timber trestles in a site yard, its links tied to the main bars
Reading time
4 minutes
Standard
EN 1992-1-1
Clauses
6.2.1, 6.2.2, 6.2.3, 9.2.2
Calculator
Beam
Updated
On this page 6 sections
  1. Step 1: The design shear
  2. Step 2: Resistance without links
  3. Step 3: The strut angle and the crushing limit
  4. Step 4: The links
  5. How this differs from BS 8110
  6. Points to remember

Shear in a reinforced concrete beam is checked in two stages in EN 1992-1-1. First, can the concrete carry the shear on its own? If not, links are designed by the variable strut inclination method, a truss in which the concrete forms inclined struts and the links form the ties.

This guide continues the beam bending example: a 300 × 500 mm beam spanning 6000 mm, with a design load of 43.313 kN/m.

The answer, worked out below

Design shear at d
110.01 kN
VEd, at d from the face of the support
Without links
82.76 kN
VRd,c, less than VEd: links are needed
Strut limit
384.42 kN
VRd,max with cot θ = 2.5
Links provided
H8 @ 325, 2 legs
VRd,s = 139.20 kN

The whole check is one decision followed by the truss:

Shear design of a beam to EN 1992-1-1
  1. Design shear VEd at the face, and at d from it
  2. Resistance without links VRd,c (Expressions 6.2.a and 6.2.b)
  3. Is VEd ≤ VRd,c?

    Yes

    1. Minimum links only (9.2.2)

    No

    1. Design the links as the ties of a truss
    2. Flattest strut the web can carry, cot θ from 2.5 down to 1
    3. Asw/s = VEd / (z fywd cot θ)
  4. Provide at least the minimum links, ρw,min of 9.2.2(5)
  5. Spacing along the beam within sl,max = 0.75d
  6. Web crushing at the face of the support (6.5)

Step 1: The design shear

The shear is largest at the supports. The calculator takes it at the face of the support, and, for a beam carrying mainly a uniform load, again at a distance from the face, where clause 6.2.1(8) lets the check for the links be made:

EN 1992-1-1 §6.2.1(1)

Design shear at the right face of support A

At the face of the support: used for the strut checks (6.9 and 6.5).

= 129.94 kN
EN 1992-1-1 §6.2.1(8)

Design shear at d from the face

For a uniformly loaded beam the links may be designed for the shear at d from the face (6.2.1(8)); this value is used for VRd,c and the links.

= 110.01 kN
VEd (kN)136.43 kN−136.43 kNABSpan 1: 6.30 mDesign value: 129.94 kN at x = 0.15 m
Shear force diagram. Every load arrangement and combination in grey; the envelopes in black and green. The links are designed for the shear at d from the face of the support.

The shear resistance of a member without shear reinforcement is given by Expression 6.2.a, with a lower limit 6.2.b:

where is the size effect (deeper members are relatively weaker in shear), is the ratio of tension steel anchored beyond the section, and in the recommended values.

EN 1992-1-1 §6.2.2(1), Eq. (6.2.a)

Size factor

d in mm.

= 1.659
EN 1992-1-1 §6.2.2(1), Eq. (6.2.a)

Tension reinforcement ratio

Asl is the tension steel extending at least (lbd + d) beyond the section (Figure 6.3).

= 0.00911
EN 1992-1-1 §6.2.2(1), Eq. (6.3N)

Minimum shear strength

= 0.410 MPa
EN 1992-1-1 §6.2.2(1), Eq. (6.2.a)
EN 1992-1-1 §6.2.2(1), Eq. (6.2.b)

Shear resistance without shear reinforcement

Expression 6.2.a governs. σcp = 0 (no axial force).

= 82.76 kN
EN 1992-1-1 §6.2.1(4)
EN 1992-1-1 §6.2.1(5)

Are designed links needed?

VEd (at d from the face of the support) exceeds VRd,c: links are designed for it (6.2.3).

Here is larger than , so designed links are needed. In a beam, clause 9.2.2 asks for at least minimum links anyway; slabs, by contrast, normally carry their shear without links.

Step 3: The strut angle and the crushing limit

In the truss model, the angle of the concrete struts may be chosen between (45°) and (about 21.8°) in the recommended values. A flatter strut means each link crosses a longer length of the crack, so fewer links are needed, but the struts are more highly stressed. The calculator tries the flattest strut first and checks that the web can carry it:

EN 1992-1-1 §6.2.2(6), Eq. (6.6N)

Strength reduction factor for concrete cracked in shear

ν₁ = ν and αcw = 1 (no prestress, no axial force).

= 0.528
EN 1992-1-1 §6.2.3(1)

Lever arm for shear

= 414.0 mm
EN 1992-1-1 §6.2.3(3), Eq. (6.9)
EN 1992-1-1 §6.2.3(2), Eq. (6.7N)

Strut resistance with cot θ = 2.5

VEd at the face is within VRd,max, so the flattest strut, cot θ = 2.5, is used (fewest links).

= 384.42 kN

is well above , so can be used. If it were not, a steeper strut would be tried.

The links needed follow from the truss (Expression 6.8):

and they may not be less than the minimum of 9.2.2(5), , nor further apart than along the beam:

EN 1992-1-1 §6.2.3(3), Eq. (6.8)

Links required for VEd

From (6.8) with VRd,s = VEd; VEd is taken at d from the face of the support (6.2.1(8), uniform load).

= 0.244 mm²/mm
EN 1992-1-1 §9.2.2(5), Eq. (9.5N)
EN 1992-1-1 §9.2.2(5), Eq. (9.4)

Minimum links

Vertical links: α = 90°, sin α = 1.

= 0.263 mm²/mm
EN 1992-1-1 §9.2.2(6), Eq. (9.6N)

Maximum spacing of links along the beam

= 345.0 mm
EN 1992-1-1 §9.2.2(8), Eq. (9.8N)

Spacing of the link legs across the beam

2 legs, 248 mm apart. Sri Lanka NA: no tension bar more than 150 mm from a leg — the farthest bar is 87 mm from one.

= 400.0 mm ✓ OK
EN 1992-1-1 §9.2.2(5)
EN 1992-1-1 §9.2.2(6)

Links provided

Spacing: sreq rounded down to 25 mm, not more than sl,max = 345.0 mm.

= 0.309 mm²/mm ✓ OK

Here the minimum governs, which is common for lightly loaded beams. The links provided give a resistance of:

EN 1992-1-1 §6.2.3(3), Eq. (6.8)

Shear resistance of the links

= 139.20 kN ✓ OK

The resistance of the links falls in proportion as they spread out. The chart runs the calculator's own link check for this beam at every spacing up to : any spacing whose curve stays above carries the shear, and the lower dashed line shows what the concrete alone could carry.

A steel fixer tying a link to a main bar of a beam cage
The links are tied to the main bars before the cage goes into the formwork; the spacing on the drawing is the s of the design.
Truss model for the links (6.2.3)compression chordtension chord (main bars)θ = 21.8°z = 414s = 325cot θ = 2.50; links H8, 2 legs. Each link carries the shear over a length z cot θ of the beam. Not to scale along the beam.
Truss model for the links. Concrete struts at θ and the links as ties; the flatter the strut (larger cot θ), the fewer links, within the limits of 6.7N.

How this differs from BS 8110

BS 8110 also starts from a concrete shear stress , but it fixes the strut at 45° and adds the concrete's share to the links. Eurocode 2 does not add the two: once links are needed, they carry the whole shear, but the flatter strut allowed by the variable angle makes them more efficient. The BS 8110 to Eurocode 2 guide lists the other differences.

Points to remember

Open the beam calculator and raise the load to see the links change, and the point at which the strut angle has to steepen.