Derive development length expression

Development length

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 $L_d=\dfrac{0.87\times415\,\phi}{4\times1.92}=47\phi$.

Check at simple support

Condition at a simple support (IS 456 cl. 26.2.3.3). At a simply supported end the moment is small, but the positive-moment bars must still be able to develop their full design stress at the sections where they are needed. To ensure this, the bar diameter is limited so that

$$L_d\le\frac{1.3\,M_1}{V}+L_0$$

Proof.

Near the support the shear $V$ is almost constant, and the bending moment grows linearly from zero at the support centre:

$$M=V\,x$$

Let $M_1$ be the moment of resistance of the section when all the tension bars at it are stressed to $0.87f_y$. The applied moment reaches $M_1$ at $x=M_1/V$ from the support centre, so the bars must be fully developed by this point.

  bar end   support centre             section where M = M1
     |<- L0 ->|<-------- M1/V -------->|
     ==========================================  bar
     |<--- available length = L0 + M1/V --->|

Available length of bar $=$ (distance from the support centre to that section) $+$ (anchorage beyond the support centre, $L_0$):

$$L_{avail}=\frac{M_1}{V}+L_0$$

The bar must satisfy $L_d\le L_{avail}$. The compressive reaction at the support confines the concrete around the bar and so increases bond. The code allows for this by increasing the first term by 30%:

$$\boxed{L_d\le\frac{1.3\,M_1}{V}+L_0}$$

where $M_1$ = moment of resistance of the section assuming all the steel at the section is stressed to $0.87f_y$; $V$ = factored shear at the section; $L_0$ = sum of the anchorage beyond the centre of the support and the equivalent anchorage value of any hook or mechanical anchorage. (At a point of inflection, $L_0$ is limited to the greater of the effective depth $d$ and $12\phi$.) The factor 1.3 is used when the bar ends are confined by a compressive reaction; otherwise $M_1/V$ is used.