--- title: "Flexural bond stress and Ld proof" subject: "Design of RCC Structure" chapter: 6 key: "semester_7/design_of_rcc_structure:flexural-bond-stress-and-ld-proof" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/flexural-bond-stress-and-ld-proof site: BCT Engineering --- # Flexural bond stress and Ld proof Design of RCC Structure · Chapter 6: Design for bond and development length - Asked in: 2075 Asoj, 7 marks - 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/flexural-bond-stress-and-ld-proof - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-6 ## Question Define anchorage bond and flexural bond stress. Prove that flexural bond stress is a function of shear force (V) and $L_d \le 1.3\frac{M_1}{V_u} + L_0$ at simple support end, where symbols have their usual meaning. ## Short answer Anchorage bond develops bar stress to avoid slip; flexural bond stress τ_bf = V/(Σ_o jd). At a simple support, M = V x; the bar must be fully developed at x = M_1/V. Available length L_0 + M_1/V, increased 30% for confinement, gives L_d ≤ 1.3 M_1/V + L_0. ## Answer #### Definitions - **Anchorage bond:** the bond developed over the embedded length of a bar that lets it carry its full design stress without slipping (developed at the ends of bars, at supports and at cut-offs). - **Flexural bond:** the bond stress caused by the change in bar force along the span as the bending moment changes (it is high near the supports, where shear is high). #### Flexural bond stress is a function of shear force **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. #### 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 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 $0.87f_y$; $V$ ($=V_u$) = 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. Why does flexural bond stress arise in a beam? A. The bars are bent into hooks at the ends B. The concrete shrinks around the bar C. The tension in the steel changes along the span as the bending moment changes D. The stirrups press the bar sideways Answer: C. T = M/jd, so a varying M gives a change in T per unit length that must be resisted by bond. (Key point: Flexural bond stress) 2. Which step shows that flexural bond stress depends on shear force? A. Σo = nπφ B. M = Vx near the support C. dT = dM/jd = V dx/jd, resisted by τbf Σo dx D. T = σs πφ²/4 Answer: C. Because dM/dx = V, the bond force per unit length equals V/jd, giving τbf = V/(Σo jd). (Key point: Link with shear) 3. In Ld ≤ 1.3M1/V + L0, what is M1? A. Moment at the support face under working load B. Moment of resistance of the concrete alone C. Moment of resistance of the section with all tension bars stressed to 0.87 fy D. Maximum service moment at mid-span Answer: C. The bars must be fully developed at the section where M1 acts, at x = M1/V from the support centre. (Key point: Meaning of M1) 4. Why is the first term multiplied by 1.3, and what is L0? A. The compressive reaction at the support improves bond; L0 is the anchorage beyond the support centre, the greater of d and 12φ B. Because shear is doubled at the support; L0 is the clear cover C. Because of a factor of safety on steel; L0 is always 12φ D. Because the bar is in compression; L0 is the lap length Answer: A. The support reaction squeezes the concrete around the bar, so the 30% increase is allowed. L0 = max(d, 12φ) beyond the centre of support. 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