--- title: "Bond and development length with derivation" subject: "Design of RCC Structure" chapter: 6 key: "semester_7/design_of_rcc_structure:bond-and-development-length-with-derivation" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/bond-and-development-length-with-derivation site: BCT Engineering --- # Bond and development length with derivation Design of RCC Structure · Chapter 6: Design for bond and development length - Asked in: 2078 Bhadra, 5 marks; 2080 Bhadra, 6 marks; 2074 Asoj (asked 3 times) - Video: [Derivation for development length ! RCC !](https://www.youtube.com/watch?v=vYC_vbV7fYo) — RS EDUCATION - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/bond-and-development-length-with-derivation - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-6 ## Question Explain anchorage bond, flexural bond stress and development length with formula derivation (derive the equation for development length). ## Short answer Anchorage bond prevents bar pull-out by developing stress over embedded length. Flexural bond stress arises from moment change, τ_bf = V/(Σ_o jd). Development length L_d is derived by equating bar force 0.87 f_y πφ²/4 to bond resistance τ_bd πφ L_d, giving L_d = 0.87 f_y φ/(4 τ_bd). ## Answer #### 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 $L_d$. ```text 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 yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the purpose of anchorage (development) bond? A. It increases the compressive strength of concrete B. It makes the shear force constant along the span C. It lets a bar reach its design stress at a section without slipping out of the concrete D. It reduces the self-weight of the beam Answer: C. Anchorage bond develops over the embedded length so the bar can carry its design stress without being pulled out. (Key point: Anchorage bond) 2. The flexural bond stress is given by τbf = V / (Σo jd). What does this show? A. It is directly proportional to the shear force V, so it is highest near supports B. It is independent of the shear force C. It is inversely proportional to V and lowest near supports D. It is proportional to the bending moment, so it is highest at mid-span Answer: A. Since dT = dM/jd = V dx/jd, the change in bar force per unit length depends on V, which is large near supports. (Key point: Flexural bond stress) 3. How is the development length equation obtained? A. Equate the bond force to the bending moment of the beam B. Equate bar tension to the concrete compressive force C. Equate πφ² Ld to the bar weight D. Equate bar tension σs(πφ²/4) to bond resistance τbd(πφ Ld) Answer: D. Equilibrium T = F gives σs πφ²/4 = τbd πφ Ld, so Ld = φσs/(4τbd) = 0.87 fy φ/(4 τbd). (Key point: Derivation of Ld) 4. For Fe415 deformed bars in M20 (plain-bar τbd = 1.2 N/mm²), what is Ld? A. About 47φ, using τbd = 1.2 × 1.6 = 1.92 N/mm² B. About 47φ, using τbd = 1.2 × 0.6 C. About 29φ, using τbd = 1.2 N/mm² increased by 25% D. About 75φ, using τbd = 1.2 N/mm² unchanged Answer: A. Deformed bars get a 60% increase in τbd, so Ld = 0.87×415φ/(4×1.92) = 47φ. 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