Anchorage Length of Reinforcement to Eurocode 2

What this check does

A bar can only carry force once it is anchored: the force has to be transferred to the concrete by bond along a length of bar. The check finds that length and compares it with what the detail provides:

$$l_{bd} \le l_{b,provided}$$

The ultimate bond stress

The design value of the ultimate bond stress for ribbed bars is:

$$f_{bd} = 2.25\,\eta_1\,\eta_2\,f_{ctd}$$

where:

• $f_{bd}$ - design ultimate bond stress (MPa)

• $\eta_1$ - bond condition coefficient: 1.0 for good bond conditions, 0.7 for all other cases

• $\eta_2$ - bar size coefficient: 1.0 for bar diameters up to 32 mm

• $f_{ctd}$ - design tensile strength of concrete, $\alpha_{ct} f_{ctk,0.05} / \gamma_c$, with $\alpha_{ct} = 1.0$ and $\gamma_c = 1.5$ recommended (MPa)

Good bond conditions depend on the bar position during concreting. Bars near the bottom of a pour are in good conditions; bars near the top of a deep pour are not, because bleed water and settlement weaken the concrete under them. EN 1992-1-1 Figure 8.2 defines the zones.

The basic required length

Balancing the bar force against bond over its perimeter gives the basic required anchorage length:

$$l_{b,rqd} = \frac{\phi}{4}\,\frac{\sigma_{sd}}{f_{bd}}$$

where:

• $l_{b,rqd}$ - basic required anchorage length (mm)

• $\phi$ - bar diameter (mm)

• $\sigma_{sd}$ - design stress in the bar at the start of the anchorage (MPa)

Using the actual stress rather than the full design yield strength is allowed, and it shortens anchorages where bars are not fully stressed.

The design length

The design anchorage length applies the alpha factors:

$$l_{bd} = \alpha_1\,\alpha_2\,\alpha_3\,\alpha_4\,\alpha_5\;l_{b,rqd} \;\ge\; l_{b,min}$$

where:

• $\alpha_1$ - shape of the bar: 1.0 for straight bars; 0.7 for bent or hooked bars in tension when the cover perpendicular to the bend is more than $3\phi$

• $\alpha_2$ - concrete cover; for straight bars in tension $1 - 0.15\,(c_d - \phi)/\phi$, between 0.7 and 1.0

• $\alpha_3$ - confinement by transverse reinforcement not welded to the main bar

• $\alpha_4$ - confinement by welded transverse bars, 0.7

• $\alpha_5$ - confinement by transverse pressure, $1 - 0.04p$

The product $\alpha_2\,\alpha_3\,\alpha_5$ may not be less than 0.7. $c_d$ is the smallest of the side cover, the bottom cover and half the clear spacing between bars. Taking every alpha as 1.0 is always conservative.

Minimum anchorage lengths

For anchorage in tension:

$$l_{b,min} = \max\{\,0.3\,l_{b,rqd};\; 10\phi;\; 100\ \text{mm}\,\}$$

For anchorage in compression:

$$l_{b,min} = \max\{\,0.6\,l_{b,rqd};\; 10\phi;\; 100\ \text{mm}\,\}$$

Bond stress by concrete class

Good bond conditions, bars up to 32 mm:

Worked example

Bar: 16 mm, B500, fully stressed, straight, in tension. Concrete: C30/37, good bond conditions. Cover: 35 mm, clear spacing between bars 60 mm.

$$f_{ctd} = \frac{1.0 \times 2.0}{1.5} = 1.33\ \text{MPa}, \qquad f_{bd} = 2.25 \times 1.0 \times 1.0 \times 1.33 = 3.00\ \text{MPa}$$

$$\sigma_{sd} = f_{yd} = \frac{500}{1.15} = 435\ \text{MPa}$$

$$l_{b,rqd} = \frac{16}{4} \times \frac{435}{3.00} = 580\ \text{mm} \approx 36\phi$$

The cover factor uses $c_d = \min\{35;\ 60/2\} = 30$ mm:

$$\alpha_2 = 1 - 0.15\,\frac{30 - 16}{16} = 0.87$$

$$l_{bd} = 0.87 \times 580 = 504\ \text{mm}$$

$$l_{b,min} = \max\{0.3 \times 580;\ 160;\ 100\} = 174\ \text{mm}$$

Result: provide at least 504 mm, rounded up to 510 mm on the drawing. The same bar in poor bond conditions ($\eta_1 = 0.7$) needs $l_{b,rqd} = 828$ mm, about $52\phi$, which is why top bars in deep beams are detailed longer.

Key points

• $f_{bd} = 2.25\,\eta_1\,\eta_2\,f_{ctd}$; poor bond conditions cut it by 30%.

• $l_{b,rqd} = (\phi/4)(\sigma_{sd}/f_{bd})$, using the actual bar stress where it is lower.

• The alpha factors can only shorten the length, and $\alpha_2\alpha_3\alpha_5 \ge 0.7$.

• Always check $l_{b,min}$, and use the larger value.

• Lap lengths build on $l_{b,rqd}$ with their own factor $\alpha_6$.