--- title: "Derive development length expression" subject: "Design of RCC Structure" chapter: 6 key: "semester_7/design_of_rcc_structure:derive-development-length-expression" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/derive-development-length-expression site: BCT Engineering --- # Derive development length expression Design of RCC Structure · Chapter 6: Design for bond and development length - Asked in: 2082 Baisakh, 4 marks; 2081 Bhadra, 3 marks; 2076 Asoj, 6 marks; 2072 Kartik, 4 marks; 2065 Kartik (old course), 10 marks (asked 5 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/derive-development-length-expression - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-6 ## Question Derive the expression $L_d \le 1.3\frac{M_1}{V} + L_0$ for development length (at a simply supported end), where the symbols have their usual meaning. ## Short answer Development length Ld = φσs/4τbd from bond equilibrium. At simple support, moment grows as M = Vx; bars must develop full stress by x = M1/V from support centre. Available length = M1/V + L0. Compressive reaction increases bond by 30%, giving code requirement Ld ≤ 1.3M1/V + L0. ## Answer #### 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 $L_d$. ```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 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What equilibrium gives Ld? A. Tension = tau_bd phi Ld B. Bar tension sigma_s pi phi^2/4 = bond force tau_bd pi phi Ld C. Compression = shear D. Moment of bar = bond moment Answer: B. This gives Ld = phi sigma_s/(4 tau_bd) = 0.87 fy phi/(4 tau_bd). (Key point: Basic equilibrium) 2. In M = Vx near the support, what is M1/V? A. Span length B. Lever arm C. Distance from support centre to section where bars reach 0.87fy D. Anchorage length Answer: C. Near the support V is nearly constant, so the moment grows linearly. (Key point: Meaning of M1/V) 3. Why is M1/V increased by 30%? A. Safety factor for steel B. Hooks C. Shear reinforcement D. The compressive support reaction confines the bar and improves bond Answer: D. This is used when the end is confined by a compressive reaction. (Key point: Why 1.3) 4. What is L0? A. Anchorage beyond the support centre, the greater of d and 12 phi B. Clear span C. Development length D. Cover Answer: A. Hooks or bends may be included. (Key point: Meaning of L0) ## Close variants set in other exams - [Define Ld and lap splice, derive Ld](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adefine-ld-and-lap-splice-derive) — Design of RCC Structure, chapter 6, asked 6× ## Related in Design of RCC Structure - [Design simply supported beam, 6 m span](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adesign-simply-supported-beam-6-m) — Design of RCC Structure, chapter 4 - [Two-way slab simply supported, 6x4 m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Atwo-way-slab-simply-supported-6x4) — Design of RCC Structure, chapter 9 - [Bond and development length with derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Abond-and-development-length-with-derivation) — Design of RCC Structure, chapter 6, asked 3× - [Define development length and lap splice](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adefine-development-length-and-lap-splice) — Design of RCC Structure, chapter 6 - [Define development length (short)](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adefine-development-length-short) — Design of RCC Structure, chapter 6 - [Development length and need of splices](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Adevelopment-length-and-need-of-splices) — Design of RCC Structure, chapter 6 ## Related in other subjects - [Lateral vibration of a simply supported beam](https://www.bctengineer.com/question.md?key=semester_7%2Ftheory_of_mechanism_and_machine_ii%3Abeam-lateral-vibration-simply-supported) — Theory of Mechanism and Machine II, chapter 10 - [Simply supported I-beam: stress at flange-web junction](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ai-beam-flange-web-junction-stress) — Strength of Materials, chapter 7 - [Castigliano: central deflection of simply supported beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acastigliano-central-deflection-ss-beam-udl) — Theory of Structures II, chapter 2 - [Simply supported T-beam: max bending stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3At-beam-6m-100kn-10kn) — Strength of Materials, chapter 7