Design for bond and development length

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 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 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, $T=F$:

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 M20, M25, M30, M35 and M40 in tension; 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 positive-moment bars must be anchored so that they can develop their full stress at the section of maximum moment, even if the moment there is not much. The code gives

Proof.

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

Let $M_1$ be the moment of resistance of the section when all the tension bars at it are stressed to $0.87f_y$. That section is at $x=M_1/V$ from the support centre; the bars must be developed from this point.

   support
  ___|___________ M1 section
  |<-- L0 -->|<-- M1/V -->|
  bar ================>
  |<-------- available length -------->|

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

The bar must satisfy $L_d\le L_{avail}$. The compressive reaction at the support squeezes the concrete around the bar and so increases bond, which is allowed for by increasing the first term by 30%:

where $M_1$ = moment of resistance of the section assuming all steel at the section stressed to $0.87f_y$; $V$ = factored shear at the section; $L_0$ = sum of the anchorage beyond the centre of support, equal to the greater of the effective depth $d$ and $12\phi$ (anchorage value of hooks or bends may be included). When the end is confined by a compressive reaction, $1.3M_1/V$ is used; otherwise $M_1/V$.

Definitions

• Development length ($L_d$): the length of bar embedded in concrete required to develop the design stress in the bar through bond: $L_d=\dfrac{\phi\,0.87f_y}{4\tau_{bd}}$.
• Lap splice: the joint of two bars made by overlapping them side by side for a specified length, so that force transfers from one bar to the other through the surrounding concrete. Lap length is at least $L_d$ (and $\ge30\phi$ for flexural tension, $2L_d$ for direct tension, $24\phi$ for compression).

Derivation of $L_d\le 1.3\dfrac{M_1}{V_u}+L_0$

Near a simple support, shear is nearly constant, so $M=V_ux$. The bar must be fully developed at the section where the moment of resistance of the bars, $M_1$ (all steel stressed to $0.87f_y$), occurs, which is at $x=M_1/V_u$ from the support centre. The length of bar available for development is the distance to that section plus the anchorage $L_0$ beyond the support centre:

For the bar to be safe against pull-out, $L_d\le L_{avail}$. The compressive reaction at the support confines the bar and enhances bond by about 30%, so the first term is multiplied by 1.3:

Here $M_1$ is the moment of resistance of the section assuming all bars stressed to $0.87f_y$, $V_u$ the factored shear at the section, and $L_0$ the greater of $d$ and $12\phi$ beyond the centre of support (including hook/bend anchorage values).

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 sufficient straight length ($L_d$), 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. 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

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

Hence flexural bond stress is directly proportional to the shear force $V$, and it is highest near supports where $V$ is large. Here $\Sigma o$ is the sum of the perimeters of the tension bars ($\Sigma o=n\pi\phi$) and $jd\approx0.9d$. The shear force $V$ is high near supports, so flexural bond stress 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 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 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, $T=F$:

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 M20, M25, M30, M35 and M40 in tension; 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 $L_d=\dfrac{0.87\times415\,\phi}{4\times1.92}=47\phi$.

Definitions

• Anchorage bond: the bond developed over the embedded length of a bar that enables it to carry its full design stress without slipping (developed at the ends of bars, at supports and at cut-offs).
• Flexural bond: bond stress due to change in bar force along the span with change in bending moment (it is high near supports where shear is high).

Flexural bond stress is a function of shear force

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

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

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

Proof of the condition at simple support

Condition at a simple support (IS 456 cl. 26.2.3.3). At a simply supported end, the positive-moment bars must be anchored so that they can develop their full stress at the section of maximum moment, even if the moment there is not much. The code gives

Proof.

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

Let $M_1$ be the moment of resistance of the section when all the tension bars at it are stressed to $0.87f_y$. That section is at $x=M_1/V$ from the support centre; the bars must be developed from this point.

   support
  ___|___________ M1 section
  |<-- L0 -->|<-- M1/V -->|
  bar ================>
  |<-------- available length -------->|

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

The bar must satisfy $L_d\le L_{avail}$. The compressive reaction at the support squeezes the concrete around the bar and so increases bond, which is allowed for by increasing the first term by 30%:

where $M_1$ = moment of resistance of the section assuming all steel at the section stressed to $0.87f_y$; $V$ = factored shear at the section; $L_0$ = sum of the anchorage beyond the centre of support, equal to the greater of the effective depth $d$ and $12\phi$ (anchorage value of hooks or bends may be included). When the end is confined by a compressive reaction, $1.3M_1/V$ is used; otherwise $M_1/V$.

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 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 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, $T=F$:

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 M20, M25, M30, M35 and M40 in tension; 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 $L_d=\dfrac{0.87\times415\,\phi}{4\times1.92}=47\phi$.

Use of development length in RCC construction

1. Anchorage of bars: bars are extended by $L_d$ beyond the point where full stress is required (beyond column face into beam, into supports, into footings).
2. Lap splices: the lap length is based on $L_d$ ($\ge L_d$ and $\ge 30\phi$ in flexure, $2L_d$ in direct tension).
3. Curtailment of bars: bars are extended by $d$ or $12\phi$ beyond the theoretical cut-off point, and the development length from the maximum-stress section is checked.
4. Support anchorage: at simple supports $L_d\le 1.3M_1/V+L_0$ is checked.
5. Column-footing and beam-column joints: dowels and column bars must be anchored by $L_d$ (for seismic work measured from the critical section).
6. Hooks and bends: the anchorage value of a standard hook is $16\phi$ and of a 90$^\circ$ bend $8\phi$, which reduces the straight length required.

Development length depends on bar diameter, steel grade, concrete grade, bar surface (plain or deformed), and tension or compression.

Proof of $L_d=\dfrac{0.87f_y\,\phi}{4\,\tau_{bd}}$

Tension in a bar of diameter $\phi$ at design stress $0.87f_y$:

Resisting bond force along embedded length $L_d$ at average design bond stress $\tau_{bd}$ on the surface area $\pi\phi L_d$:

Equating $T=F$:

Proof of $S_v=\dfrac{0.87f_yA_{sv}d}{V_s}$

Consider a diagonal crack at 45$^\circ$ in a beam of effective depth $d$. The horizontal length of the crack is about $d$ (for a vertical stirrup the crack crosses $d/s_v$ stirrups).

        crack at 45 deg
   ____________/_________
  |           /  |  |  |
  |          /   |  |  |  stirrups, spacing Sv
  |_________/____|__|__|
  |<--- d --->|

Number of stirrups crossing the crack $=\dfrac{d}{S_v}$. Each stirrup (two legs, total area $A_{sv}$) carries design tension $0.87f_yA_{sv}$ at yield. The vertical component of the force carried by the stirrups equals the shear $V_s$ assigned to them (vertical stirrups):

where $V_s=V_u-\tau_cbd$ is the shear to be carried by the stirrups.

• Development length ($L_d$): the length of reinforcing bar that must be embedded in concrete beyond a section so that the bar can develop its full design stress ($0.87f_y$) through bond without slipping. $L_d=\dfrac{0.87f_y\,\phi}{4\tau_{bd}}$, e.g. about $47\phi$ for Fe415 deformed bars in M20.
• Lap splice: a joint between two reinforcing bars made by placing them side by side and overlapping them for a specified length, so that the force passes from one bar to the other through the concrete by bond. The lap length is not less than $L_d$ (or $30\phi$ for bars in flexural tension, $24\phi$ in compression, $2L_d$ or $30\phi$ in direct tension). Laps should be staggered and avoided at sections of maximum stress.

Development length is the length of a reinforcing bar, embedded in concrete, that is required to transfer the design stress in the bar to the surrounding concrete by bond, so that the bar will not pull out. It is given by $L_d=\dfrac{0.87f_y\,\phi}{4\,\tau_{bd}}$ (IS 456 cl. 26.2.1), where $\phi$ is the bar diameter and $\tau_{bd}$ the design bond stress.

Development length

Development length $L_d$ is the length of embedded bar needed to develop its design stress $0.87f_y$ by bond with concrete:

For example, for Fe415 deformed bars in M20 concrete, $\tau_{bd}=1.2\times1.6=1.92$ N/mm$^2$ and $L_d=47\phi$.

Why splices are required

Reinforcing bars are available only in standard lengths (usually 12 m, less in practice because of transport). Splices (joints) are therefore needed:

1. Limited length of bars: structures longer than the available bar length, such as long beams, tall columns and walls, need bars to be joined.
2. Transport and handling: long bars are difficult to carry, store and fix on site.
3. Construction stages: column bars are projected upward at each floor, and lapped with next-lift bars (construction joints).
4. Continuity of load path: splices transfer the full force from one bar to the next, so reinforcement acts as a continuous bar.
5. Economy: short offcuts of bars can be used instead of wasting them.

Types: lap splices (most common), welded joints and mechanical couplers. Laps should be staggered, be at least $L_d$ or $30\phi$ (flexural tension), and should not be at sections of maximum stress.