Shear and links

Why beams crack at an angle near their supports, how links stop that, and how to choose and space them to SANS 10100-1. Two worked examples, by hand and in STATIX.

1. Learning outcomes

After this lesson you can:

explain what shear does to a beam and why links are needed;
find the design shear force at the critical section and the shear stress v = V/(b d);
check v against the maximum, and work out v_c, the shear stress the concrete carries;
decide between nominal and designed links and work out the link spacing;
lay out the links along a beam in zones.

2. Before you start

Lesson 1 and lesson 2: this lesson uses both of their beams.
The free STATIX RC beam calculator, open in a second tab.
A pocket calculator, paper and a pencil.
About 60 minutes.

3. What shear does to a beam

Bending is not the only thing a beam has to resist. The shear force V tries to slide one part of the beam down past the next. Shear is largest at the supports, where bending is smallest.

Shear and bending together pull the concrete apart on a slope of about 45°. Concrete is weak in tension, so once the pull is large enough, inclined cracks form near the supports.

What you would see in a test. Near mid-span the bending moment is large and the cracks are vertical. Near the supports the shear force is large and the cracks lean at about 45°. Links are there to stop these inclined cracks from opening.

An inclined crack is dangerous because it runs across the bottom bars, not along them. The bottom bars cannot stop it from opening. Something must cross the crack the other way: that is the job of the links.

4. How a cracked beam carries shear

Engineers picture a cracked beam as a truss. The picture below builds it up in three steps. Read it from top to bottom.

How a cracked beam carries shear, built up in three steps (left half of the beam). The concrete between the inclined cracks works as sloping struts in compression. Each strut pushes down and outwards; a link at its top end hangs it back up to the top of the beam. Together with the concrete at the top and the bottom bars, the struts and links form a truss. Links are the vertical members of that truss.

The truss gives two simple rules. First, the links are in tension, so we need enough link steel. Second, the concrete struts are in compression, so the concrete itself has a limit: if the shear is too large, the struts crush, and no amount of links will help.

5. The rules (SANS 10100-1, 4.3.4)

The shear stress v

v = V / (b d) in MPa, with V in N and b, d in mm

v is an average shear stress over the web. It is the number we compare with the limits below.

The upper limit

v ≤ 0.75 √f_cu and v ≤ 4.75 MPa f_cu = 30 MPa: 0.75 × √30 = 4.11 MPa

This is the crushing limit of the struts. Check it with V at the face of the support. If v is higher, make the section wider or deeper.

The concrete's share v_c

v_c = (0.75/1.4) × (f_cu/25)^1/3 × (100A_s/(b d))^1/3 × (400/d)^1/4

A_s is the tension steel that carries on past the section (well anchored). Use 100A_s/(b d) up to 3 and f_cu up to 40 MPa. Example B below shows what each of the four numbers means.

Which links?

v ≤ v_c + 0.4 → nominal links: A_sv/s_v ≥ 0.4 b / (0.87 f_yv) v > v_c + 0.4 → designed links: A_sv/s_v ≥ b (v − v_c) / (0.87 f_yv) Spacing along the beam: s_v ≤ 0.75d

A_sv is the area of all the legs of one link (two legs for a normal link), s_v is the spacing along the beam, and f_yv is the yield strength of the link steel. Nominal links are a minimum: in beams they go in everywhere, even where v is below v_c.

Where to check

For a beam with a uniform load, design the links for the shear at a distance d from the face of the support. Load closer to the support than d goes straight down into the support through a strut. In this lesson the supports are narrow, so we measure d from the support line, which is slightly on the safe side.