Lecture 4: Brackets and Corbels Design (ACI Code) PDF

# Libyan Academy for Postgraduate Studies ## Department of Civil and Architectural Engineering/Structural Engineering ### Course: Advanced Reinforced Concrete Design- A(CES605) ## Lecture (4): Brackets and Corbels Design Based on ACI Code ### 1- Introduction: Brackets and corbels are short cantile...

# Libyan Academy for Postgraduate Studies ## Department of Civil and Architectural Engineering/Structural Engineering ### Course: Advanced Reinforced Concrete Design- A(CES605) ## Lecture (4): Brackets and Corbels Design Based on ACI Code ### 1- Introduction: Brackets and corbels are short cantilever beams that project from columns or walls to support loads, such as beams or slabs. Their structural behavior differs from that of conventional beams due to their short span, large, applied forces (high shear stresses), and deep section relative to their length. The ACI 318 code provides guidelines for the design of these elements, which are critical in ensuring the stability and performance of the supported structure. - **Brackets:** Short cantilevers used to support concentrated loads, usually projecting from columns or walls. - **Corbels:** Short cantilevers that extend from columns or walls, typically deeper in section, used to transfer loads from horizontal members (such as beams) to vertical members (like columns). ### 2- Failures Mode: | Mode of Failure | Causes | |---|---| | 1-Flexture Failure | 1-Small amount of main steel. <br> 2- av/d ≤1 and if cantilever fixture not clearly considered | | 2-Diagonal Splitting | 1-Low compressive strength with high main steel.<br> 2-Small amount of steel crossing the strut. | | 3-Shear Failure | 1-Small amount of stirrups.<br> 2-av/d ≤1 and if shear behaviours not clearly considered | | | **Type 1:** 1-Loss of anchorage.<br> 2-Lack in the detailing of embedded length. | | | **Type 2:** 1-Small thickness of corbel.<br>2-High value of horizontal load. | | | **Type 3:** 1-Lack in the seat plate detailing. | ### 3- Behavior of Brackets and Corbels Brackets and corbels experience high shear forces and moments at their connection to the supporting member. Due to the short projection, they behave differently from typical beams: - **Shear-controlled design:** The shear force often governs the design rather than flexure. - **Strut-and-tie models (STM):** This is frequently used for analysis and design since **it** better captures the load transfer mechanism. - **Horizontal tension reinforcement:** Required to resist the horizontal forces that arise due to the bending moments and to provide additional capacity for crack control. ### 4- Design Considerations Bracket may be considered as a very short, cantilevered beam, with flexural tension at the column face resisted by the top bars Asc. Moreover, the structural performance of a bracket can be visualized easily by means of the strut-and-tie model shown below. The provisions of ACI Code for the design of brackets and corbels is considered that: - **span ratio. <br> av/d≤1 Apply flexural model <br> a/d ≤2 Apply strut-and-tie models** - **depth at the outside edge of the bearing area** - **The usual design basis** t ≥ 0.5d <br> Mu ≤ Ø Mn and Vu ≤ Ø Vn where Ø = 0.75 - In the absence of a roller or low-friction support pad, **a** horizontal tensile force must assume to be Nue ≥ 0.2 Vu - **To find d:** Vn ≥ Vu/ Ø <br> $V_n = \dfrac{0.2 fc bd}{(3.3+0.08 fc) bd}$ (Normal weight concrete) <br> $V_n = \dfrac{11 bd}{(0.2-0.07a/d)fb'd}$ (Lightweight concrete) <br> $V_n = \dfrac{(5.5+1.9 av/d) bd}{}$. - **Mu at section at the face of the supporting:** $Mu = V_u * av + Nuc(h-d)$. ### The amount of steel Af to resist the moment Mu: $A_f = \dfrac{Mu}{∅f_y(d-a/2}$ <br> $a = \dfrac{A_ff_y}{0.85fb}$ <br> **Area of steel Af for nominal tensile strength:** $A_n = \dfrac{Nuc}{∅f_y}$ <br> - **Shear-friction reinforcement Avf:** $A_{vf} = \dfrac{V_u}{φμεξ_v}$ <br> friction factor (μ = 1.4 λ) <br> λ = 1 for normal weight concrete <br> λ= 0.75 for all-light weight concrete **The total area required for flexure and direct tension at the top of the bracket is thus:** $A_sc ≥ \dfrac{A_f+ A_n}{2}$ <br> $A_sc ≥ A_{ur} + A_n$. **Note that:** Always Asc is not less than $A_sc ≥ 0.04(f_c/f_y)bd$. - **Closed hoop stirrups having area Ah:** $A_h ≥ 0.5(A_sc -A_n)$ and $A_h ≥ 0.5 A_f$ and $A_h ≥ A_{vf}/3$. **Example:** A column bracket having the general features shown below is to be designed to carry the end reaction from a long-span precast girder. Vertical reactions from service dead and live loads are 110 and 225 kN, respectively, applied at av = 140 mm from the column face. A steel bearing plate will be provided for the girder, which will rest directly on a 125 X 75 X 10 mm steel angle anchored at the outer corner of the bracket. Bracket reinforcement will include main steel Ase welded to the underside of the steel angle, closed hoop stirrups having total area Ah distributed appropriately through the bracket depth, and framing bars in a vertical plane near the outer face. Select appropriate concrete dimensions, and design reinforcement. Material strengths are f'c = 35 MPa and f"y = 420 MPa. **Ans**: $V_u = W_u = 1.2D + 1.6L $ $ = 1.2 × 110 + 1.6 × 225 = 492 KW$ <br> $V_u = 492 KW$ - **Paveller or low-friction Support Pad.** $N_{uc} = 0.2 V_u = 0.2 × 492 = 98 KW$ **To find d:** Normal shear strength to $V_u$ must not exceed $V_n$: $V_n = (3.3 + 0.08f’_cbd)$ $4.92 × 10^3 = 0.75 × 6.1 × 300 d$ : d = 35.8 mm ~ d = 375 mm **Cover + Sars = 25 mm** $t_1, h = 375 + 25 = 400 mm$ - **Total Shear friction steel (Avf)** $A_{vf} = \dfrac{Vu }{φμξ_v}$= $ \dfrac{492 × 10^3}{0.75 × 1.4 × 1.8 × 420} $= 1116 mm² $A_{vf} = 1116 mm²$ - **find An for nominal tensile strenght** $A_n = \dfrac{N_{uC}}{∅f_y} = \dfrac{98 × 10^3}{0.75 × 420} = 311 mm² $ $A_n = 311 mm²$ - **find total rebar area at the top of the bracket** $A_{sc} = A_f + A_n = 626 + 311 = 932 mm²$ and $A_{sc} ≥ \dfrac{A_{uf}+A_n}{2} = \dfrac{1}{2} × 1116 +311 = 1055 mm²$ : Use the greatest $A_{sc} = 1055 mm²$ **Check:** $ A_sc min = 0.04 f’_c bd$ $ = 0.04 × 300 × 300 × 350 = 35010550$ - **find bending moment Mu ** $Mu = V_u * av + N_{uc} (h-d)$ $ = 492 × 0.14 + 98 × 0.25 = 71 KN.m$ Assume flexural compression stress block depth $c_a $ $c_a = 0.5 h = 50 mm$ : $A_p = \dfrac{Mu}{∅f_y(d-a/2)} = \dfrac{71×10^6}{0.25×420(375-50)} = 644 mm^2$ **Check block a: ** $a = \dfrac{A_p∅f_y}{0.85f’_cb} = \dfrac{644 × 420}{0.85 × 35 × 300} = 30 mm$ : Re find Ap for actual a = 30 $A_p = \dfrac{71×10^6}{0.75×420(375-50)} = 626 mm^2$ - **So use 22 feills to 3022 Asc=1161 mm²** - **find Ahp returns ane (closet stirrups):** $A_h = 0.5(A_{sc} - A_n)$ $ = 0.5(1161 - 311) = 425 mm²$ and $A_h = 0.5 A_f = 0.5 × 626 = 313 mm²$ and $A_h = \dfrac{1}{3} A_{vf} = \dfrac{1}{3} ×1116 = 372 mm²$ : Use $A_h = 425 mm²$ $A_h= 425 mm²$ - **say ∅10 → T 425 = 2,7 use 3 ∅10 Than → 471 mm²** - **Breaket design** b = 300 mm, d = 375 mm, h = 400 mm, t = 190 mm - **Asc = 1140.4 mm² → 3∅22** - **Ah = 471 mm² → 3∅10**