Bond and development length with derivation

Anchorage bond (development bond)

Anchorage bond is the bond stress developed over the embedded length of a bar so that the bar can reach its design stress at a section without slipping out of the concrete. It is provided by enough straight length ($L_d$), or by hooks, bends or mechanical anchorage at the ends, e.g. at simple supports, in beam-column joints and at cut-off points.

Flexural bond stress

Flexural bond stress is the bond stress that arises because the force in the tension steel changes along the span as the bending moment changes.

Derivation. For a beam section, the tension in the steel is $T=\dfrac{M}{jd}$, where $jd$ is the lever arm. Over a small length $dx$, the change in tension is

$$dT=\frac{dM}{jd}=\frac{V\,dx}{jd}$$

This change is resisted by bond over the perimeter of the bars, $\Sigma o$, along $dx$:

$$\tau_{bf}\,\Sigma o\,dx=\frac{V\,dx}{jd}\ \Rightarrow\ \boxed{\tau_{bf}=\frac{V}{\Sigma o\ jd}}$$

Here $\Sigma o$ is the sum of the perimeters of the tension bars ($\Sigma o=n\pi\phi$) and $jd\approx0.9d$. Flexural bond stress is therefore directly proportional to the shear force $V$. It is highest near the supports, where $V$ is large, so it is checked there.

Development length and its derivation

Development length $L_d$ is the length of bar needed on either side of a section so that the stress $\sigma_s$ in the bar can be developed by bond with the concrete (IS 456 cl. 26.2.1).

Derivation. Consider a bar of diameter $\phi$ stressed to $\sigma_s=0.87f_y$ at a section, anchored over a length $L_d$.

  concrete
 +---------------------------+
 |  T = σs.(πφ²/4) <==[bar]=====>|
 +---------------------------+
      |<------- Ld -------->|   bond stress τbd on πφ.Ld

Tensile force in the bar: $T=\sigma_s\times\dfrac{\pi\phi^2}{4}$.

Resisting bond force over the embedded length: $F=\tau_{bd}\times\pi\phi\,L_d$.

For equilibrium, $T=F$:

$$\sigma_s\frac{\pi\phi^2}{4}=\tau_{bd}\,\pi\phi\,L_d\ \Rightarrow\ L_d=\frac{\phi\,\sigma_s}{4\,\tau_{bd}}=\frac{0.87f_y\,\phi}{4\,\tau_{bd}}$$

Here $\tau_{bd}$ is the design bond stress (IS 456 cl. 26.2.1.1): 1.2, 1.4, 1.5, 1.7 and 1.9 N/mm$^2$ for plain bars in tension in M20, M25, M30, M35 and M40. It is increased by 60% for deformed bars and by 25% for bars in compression. For Fe415 deformed bars in M20: $\tau_{bd}=1.2\times1.6=1.92$ N/mm$^2$, so $L_d=\dfrac{0.87\times415\,\phi}{4\times1.92}=47\phi$.