CIVL 737 Design of Concrete Structures with Fibre-Reinforced Polymers (FRP) Section 2 – Reinforcing Concrete with FRP Composites Dr. Ahmed Bediwy, PEng Assistant Professor, Dept. of Civil Engineering Office: F1-1087 Lakehead University Section 2 – Reinforcing Concrete with FRP Composites Outlines 1. Constituents and Properties 2. Field Applications 3. Flexural Design of FRP-RC Elements 4. Shear Design of FRP-RC Elements 2 Lakehead University 2.3 Flexural Design of FRP-RC Elements Introduction • FRP Composite bars have high strength (800 – 2000 MPa) but low (~ 60 GPa) and medium (130 GPa) modulus of elasticity compared to steel. • The design of concrete elements reinforced with FRP bars is governed by the Serviceability Requirements: deflection and crack width. • FRP-reinforced concrete elements should be designed as over-reinforced sections to avoid the brittle rupture of FRP bars. • Material resistance factors (ϕ) similar to those used for conventional steel-reinforced concrete elements taking into account durability issues. 3 Lakehead University 2.3 Flexural Design of FRP-RC Elements Design Assumptions (Clause 8.4.1) • Strain in reinforcement and concrete shall be assumed to be directly proportional to the distance from the neutral axis (linear strain distribution). • FRP bond to concrete is equivalent to steel reinforcement (perfect bond). • The ultimate strain at the extreme concrete compression fibre (concrete crushing strain) shall be assumed to be 0.0035. • The tensile strength of concrete shall be neglected. • The compressive strength of FRP reinforcement shall be disregarded. • The stress-strain relationship of FRP is linear–elastic until failure (No yielding plateau). 4 Lakehead University 2.3 Flexural Design of FRP-RC Elements Design Process • Based on the ultimate limit state (ULS) design philosophy (Clause 6.2.3): ϕ R n > α Sn ϕ is the material resistance factor Rn is the nominal resistance α is the load factor Sn is the nominal load effect • Material resistance factors for FRP bars: Code CSA S806 (Buildings) (Clause 7.1.6.3): ϕ = 0.75 (GFRP, CFRP & AFRP) Code CSA S6 (Bridges) (Clause 16.5.6): ϕ = 0.65 (GFRP), = 0.65 (AFRP), = 0.80 (CFRP) 5 Lakehead University 2.3 Flexural Design of FRP-RC Elements Modes of Failure Mode of Failure Behaviour of Element Balanced CompressionControlled FRP Rupture Simultaneous FRP Rupture and Concrete Crushing Concrete Crushing Least Desirable: rupture is sudden and violent Hypothetical Condition Most Desirable: sufficient warning Reinforcement Ratio ρfrp < ρbal ρfrp = ρbal ρfrp > ρbal Strains εfrp = εfrpu εc < εcu εfrp = εfrpu εc = εcu εfrp < εfrpu εc = εcu Desirability 6 TensionControlled Lakehead University 2.3 Flexural Design of FRP-RC Elements Balanced Condition b cu cb d Afrpb frpu Cross Section Strain Distribution Objective: Solve for balanced reinforcement ratio (ρfrpb) Step 1: Strains Concrete: εc = εcu = 0.0035 FRP: εfrp = εfrpu = ffrpu/Efrp 7 Lakehead University 2.3 Flexural Design of FRP-RC Elements Balanced Condition b cu cb d Afrpb frpu Cross Section Strain Distribution Step 2: Strain Compatibility 𝑐𝑏 ε𝑐𝑢 0.0035 = = 𝑑 ε𝑐𝑢 + ε𝑓𝑟𝑝𝑢 0.0035 + ε𝑓𝑟𝑝𝑢 8 Lakehead University 2.3 Flexural Design of FRP-RC Elements Balanced Condition b 1Φcf’c cu a = 1c cb d Afrpb frpu Cross Section Strain Distribution Φfrpffrpu T Stress Distribution Equiv. Stress Distribution Step 3: Stress Distribution Actual stress block is non-linear based on concrete properties An equivalent rectangular stress block is used instead 𝛼1 = 0.85 − 0.0015𝑓𝑐′ ≥ 0.67 𝛽1 = 0.97 − 0.0025𝑓𝑐′ ≥ 0.67 9 Lakehead University 2.3 Flexural Design of FRP-RC Elements Balanced Condition b 1Φcf’c cu a = 1c cb d C Afrpb Φfrpffrpu T Stress Distribution Equiv. Stress Distribution frpu Cross Section Strain Distribution Step 4: Equilibrium 𝐶=𝑇 𝑇 = 𝐴𝑓𝑟𝑝𝑏 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 10 𝐶 = 𝛼1 ϕ𝑐 𝑓𝑐′ 𝛽1 𝑐𝑏 . 𝑏 Lakehead University 2.3 Flexural Design of FRP-RC Elements Balanced Condition b 1Φcf’c cu a = 1c cb d C Afrpb frpu Cross Section Strain Distribution Φfrpffrpu T Stress Distribution Equiv. Stress Distribution Step 5: Obtain Balanced Reinforcement Ratio 𝛼1 ϕ𝑐 𝑓𝑐′ (𝛽1 𝑐𝑏 . 𝑏) 𝛼1 ϕ𝑐 𝑓𝑐′ (𝛽1 . 𝑏) 0.0035 𝐴𝑓𝑟𝑝𝑏 = = 𝑑 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 0.0035 + ε𝑓𝑟𝑝𝑢 𝐴𝑓𝑟𝑝𝑏 𝛼1 𝛽1 ϕ𝑐 𝑓𝑐′ 0.0035 ρ𝑓𝑟𝑝𝑏 = = 𝑏𝑑 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 0.0035 + ε𝑓𝑟𝑝𝑢 11 Lakehead University 2.3 Flexural Design of FRP-RC Elements Compression-Controlled Failure b cu c d Afrp>Afrpb frp Cross Section Strain Distribution Objective: Solve for moment resistance (Mr) Step 1: Strains Concrete: εc = εcu = 0.0035 FRP: εfrp < εfrpu 12 Lakehead University 2.3 Flexural Design of FRP-RC Elements Compression-Controlled Failure b 1Φcf’c cu a = 1c c C d Afrp>Afrpb Φfrpffrp T Stress Distribution Equiv. Stress Distribution frp Cross Section Strain Distribution Step 2: Equilibrium 𝐶=𝑇 𝑇 = 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝 13 𝐶 = 𝛼1 ϕ𝑐 𝑓𝑐′ 𝛽1 𝑐. 𝑏 Lakehead University 2.3 Flexural Design of FRP-RC Elements Compression-Controlled Failure b 1Φcf’c cu a = 1c c C d Afrp>Afrpb Φfrpffrp T Stress Distribution Equiv. Stress Distribution frp Cross Section Strain Distribution Step 3: Rearrange and solve for 𝛽1 𝑐 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝 𝛽1 𝑐 = 𝛼1 ϕ𝑐 𝑓𝑐′ 𝑏 𝑓𝑓𝑟𝑝 = ε𝑓𝑟𝑝 𝐸𝑓𝑟𝑝 14 Lakehead University 2.3 Flexural Design of FRP-RC Elements Compression-Controlled Failure b 1Φcf’c cu a = 1c c C d Afrp>Afrpb Φfrpffrp T Stress Distribution Equiv. Stress Distribution frp Cross Section Strain Distribution Step 4: Use strain compatibility to obtain a second equation ε𝑓𝑟𝑝 ε𝑐𝑢 15 = 𝑑−𝑐 𝑐 → ε𝑓𝑟𝑝 = ε𝑐𝑢 𝑑−𝑐 𝑐 → 𝑓𝑓𝑟𝑝 = 𝐸𝑓𝑟𝑝 ε𝑐𝑢 𝑑−𝑐 𝑐 Lakehead University 2.3 Flexural Design of FRP-RC Elements Compression-Controlled Failure b 1Φcf’c cu a = 1c c C d Afrp>Afrpb frp Cross Section Strain Distribution Φfrpffrp T Stress Distribution Equiv. Stress Distribution Step 5: Solve the two equations together 𝑓𝑓𝑟𝑝 = 0.5𝐸𝑓𝑟𝑝 ε𝑐𝑢 16 4𝛼1 𝛽1 ϕ𝑐 𝑓𝑐′ 1+ ρ𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝐸𝑓𝑟𝑝 ε𝑐𝑢 1Τ 2 −1 Lakehead University 2.3 Flexural Design of FRP-RC Elements Compression-Controlled Failure b 1Φcf’c cu a = 1c c C d Afrp>Afrpb Φfrpffrp T Stress Distribution Equiv. Stress Distribution frp Cross Section Strain Distribution Step 6: Determine stress block depth, a Step 7: Solve for Mr 𝑀𝑟 = 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝 17 𝑎= 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝 𝛼1 ϕ𝑐 𝑓𝑐′ 𝑏 𝑎 𝑑− 2 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure b c Less than ultimate Ultimate c d Afrp<Afrpb ffrpu frpu Cross Section Strain Distribution Stress Distribution Objective: Solve for moment resistance (Mr) Step 1: Strains Concrete: εc < εcu FRP: εfrp = εfrpu = ffrpu/Efrp 18 Stress block parameters do NOT apply! Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure b c Less than ultimate Ultimate c d Afrp<Afrpb ffrpu frpu Cross Section Strain Distribution Stress Distribution Step 2: Assume a value for neutral axis depth, c, and obtain strain in concrete from strain distribution ε𝑓𝑟𝑝𝑢 𝑑 − 𝑐 = ε𝑐 𝑐 19 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure b c Less than ultimate Ultimate c d Afrp<Afrpb ffrpu frpu Cross Section Strain Distribution Stress Distribution Step 3: Determine modified stress block parameters 𝛼 and 𝛽 From tables and figures in ISIS Manual 3 20 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure 21 𝛼 from Figure 6.4 in ISIS Manual 3 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure 22 𝛽 from Figure 6.5 in ISIS Manual 3 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure 𝛼 and 𝛽 from Tables B.1 to B.3 in ISIS Manual 3 εo is the strain in concrete at peak stress (fc ’), which can be calculated from Table 5.7 in ISIS Manual No. 3. 23 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure b Φcf’c c a = c c C d Afrp<Afrpb frpffrpu frpu Cross Section Strain Distribution Stress Distribution Step 4: Equilibrium T Equiv. Stress Distribution Substitute assumed c 𝐶=𝑇 𝑇 = 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 𝐶 = 𝛼ϕ𝑐 𝑓𝑐′ 𝛽𝑐. 𝑏 If equilibrium is not satisfied, select a new neutral axis depth, c, and reiterate until convergence is obtained. 24 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure b Φcf’c c a = c c C d Afrp<Afrpb frpffrpu frpu Cross Section Strain Distribution Stress Distribution Step 5: Solve for Mr 𝑀𝑟 = 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 𝑑 − 25 T Equiv. Stress Distribution 𝛽𝑐 2 Lakehead University 2.3 Flexural Design of FRP-RC Elements Tension-Controlled Failure Assume c Determine 𝛼 and 𝛽 Check if equilibrium is satisfied Yes Compute Mr No Calculate new c 26 Lakehead University 2.3 Flexural Design of FRP-RC Elements Minimum Reinforcement (Clause 8.4.2 in CSA S806) 𝑀𝑟 > 1.5𝑀𝑐𝑟 𝑀𝑐𝑟 = 𝑓𝑟 𝐼𝑔 𝑦𝑡 (Clause 8.3.2.6) and 𝑓𝑟 = 0.6λ 𝑓𝑐′ (Clause 8.3.2.8) • This condition can be waived if 𝑀𝑐𝑟 > 1.5𝑀𝑓 • In slabs, a minimum area of reinforcement of (400/Ef) Ag shall be used in each of the two orthogonal directions. This reinforcement shall not be less than 0.0025 Ag and shall be spaced no farther apart than three times the slab thickness or 300 mm, whichever is less. • All FRP-RC sections shall be designed as over-reinforced sections (Clause 8.2.1). • Under-reinforced sections are allowed only when 𝑀𝑟 > 1.6𝑀𝑓 (Clause 8.2.2). 27 Lakehead University 2.3 Flexural Design of FRP-RC Elements Minimum Reinforcement (Clause 16.8.2.2 in CSA S6) 𝑀𝑟 > 1.5𝑀𝑐𝑟 𝑀𝑐𝑟 = 𝑓𝑟 𝐼𝑔 𝑓𝑟 = 0.4λ 𝑓𝑐′ (Clause 8.4.1.8.1) and 𝑦𝑡 • Under-reinforced sections are allowed only when 𝑀𝑟 > 1.5𝑀𝑓 (Clause 8.2.2). Concrete Cover • Adequate cover is required to: ✓ Protect reinforcement from fire exposure ✓ Prevent cracking due to thermal expansion Code 28 Cover Requirement CSA S806 (Clause 8.2.3) 2.0db or 30 mm CSA S6 (Clause 16.4.5) 35 ± 10 mm Lakehead University 2.3 Flexural Design of FRP-RC Elements h = 600 mm Example 1 • Calculate the moment resistance, Mr, for a rectangular section in a building with the following dimensions and properties: ✓ Clear cover to the flexural reinforcement, cc= 40 mm ✓ 8 mm-diameter CFRP bars for tension reinforcement 5 mm-diameter ✓ 5 mm-diameter CFRP bars for stirrups CFRP stirrups ✓ Concrete compressive strength, fc’ = 35 MPa. 4-8 mm diameter CFRP ✓ Flexural CFRP reinforcement properties: • Tensile strength, ffu = 2,250 MPa, b = 350 mm bars • Tensile modulus of elasticity, Ef = 147 GPa, • Ultimate tensile strain, ᵋfu = 15,300x10-6, • The area of one 8-mm diameter bar, Ab = 50 mm2. 29 Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 • Use CSA S806-12 • Calculate effective depth 8 𝑑 = 600 − 40 − = 556 𝑚𝑚 2 • Calculate the actual reinforcement ratio 𝐴𝑓𝑟𝑝 4 × 50 ρ𝑓𝑟𝑝 = = = 1.03 × 10−3 𝑏𝑑 350 × 556 • Calculate the balanced reinforcement ratio 𝛼1 = 0.85 − 0.0015𝑓𝑐′ = 0.85 − 0.0015 × 35 = 0.79 𝛽1 = 0.97 − 0.0025𝑓𝑐′ = 0.97 − 0.0025 × 35 = 0.88 ρ𝑓𝑟𝑝𝑏 = 𝛼1 𝛽1 ϕ𝑐𝑓𝑐′ 0.0035 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 0.0035+ε𝑓𝑟𝑝𝑢 10−3 ✓ Since ρ𝑓𝑟𝑝 < ρ𝑓𝑟𝑝𝑏 30 = 0.79 × 0.88 × 0.65×35 0.0035 0.75×2250 0.0035+0.0153 = 1.74 × Tension-Controlled Failure (Under-Reinforced Section) Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 31 Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 • Assume c to use strain compatibility. • A good starting point is the ratio between the actual reinforcement ratio and the balanced one. ρ𝑓𝑟𝑝 1.03 = = 0.6 ρ𝑓𝑟𝑝𝑏 1.74 • Assume 𝑐 = 0.6 𝑐𝑏 ε𝑐𝑢 0.0035 𝑐𝑏 = 𝑑= × 556 = 103.5 𝑚𝑚 ε𝑐𝑢 + ε𝑓𝑢 0.0035 + 0.0153 𝑐 = 0.6 × 103.5 = 62 𝑚𝑚 • From strain compatibility: 𝑐 62 ε𝑐 = ε𝑓𝑟𝑝𝑢 = × 0.0153 = 0.0019 𝑑−𝑐 556 − 62 32 Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 • From Fig 6.4 for concrete strength of 35 MPa: α = 0.85 33 Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 • From Fig 6.5 for concrete strength of 35 MPa: β = 0.72 34 Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 • Calculate C: 𝐶 = 𝛼ϕ𝑐 𝑓𝑐′ 𝛽𝑐. 𝑏 = 0.85 × 0.65 × 35 × 0.72 × 62 × 350 = 302.1 𝑘𝑁 • Calculate T: 𝑇 = 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 = 200 × 0.75 × 2250 = 337.5 𝑘𝑁 • 𝐶≠𝑇 Re-iterate! • Assume 𝑐 = 66 𝑚𝑚 • From strain compatibility: 𝑐 66 ε𝑐 = ε = × 0.0153 = 0.0021 𝑑 − 𝑐 𝑓𝑟𝑝𝑢 556 − 66 • From Fig 6.4 and 6.5: α = 0.89 and β = 0.72 𝐶 = 𝛼ϕ𝑐 𝑓𝑐′ 𝛽𝑐. 𝑏 = 0.89 × 0.65 × 35 × 0.72 × 66 × 350 = 336.8 𝑘𝑁 • 𝐶=𝑇 Good! 35 Lakehead University 2.3 Flexural Design of FRP-RC Elements Solution to Example 1 • Calculate moment of resistance: 0.72 × 66 𝑀𝑟 = 𝐴𝑓𝑟𝑝 ϕ𝑓𝑟𝑝 𝑓𝑓𝑟𝑝𝑢 𝑑 − = 200 × 0.75 × 2250 × 556 − 2 2 = 179.6 × 106 𝑁. 𝑚𝑚 = 179.6 𝑘𝑁. 𝑚 • Check Mr against Mcr: 𝑀𝑟 > 1.5𝑀𝑐𝑟 𝛽𝑐 𝑀𝑐𝑟 = 𝑓𝑟 𝐼𝑔 𝑦𝑡 = 0.6λ ′ 𝐼𝑔 𝑓𝑐 = 0.6 𝑦 𝑡 35 × 350×6003 12×300 = 74.5 × 106 𝑁. 𝑚𝑚 1.5𝑀𝑐𝑟 = 1.5 × 74.5 × 106 = 111.8 × 106 𝑁. 𝑚𝑚 𝑀𝑟 = 179.6 𝑘𝑁. 𝑚 > 1.5𝑀𝑐𝑟 = 111.8 𝑘𝑁. 𝑚 OK • Check Mr against Mf: Under-reinforced sections are allowed only when 𝑀𝑟 > 1.6𝑀𝑓 (Clause 8.2.2). 36 Lakehead University 2.3 Flexural Design of FRP-RC Elements Serviceability Strength • Required to resist applied loads Serviceability • Required to minimize cracking and deflection • The lower modulus of elasticity of FRP bars results in larger cracks and deflections. • In many cases, serviceability requirements govern the design of FRP-RC members. 37 Lakehead University 2.3 Flexural Design of FRP-RC Elements Serviceability (Service Stress) b N.A. d Afrp Un-cracked (Compression) c kd Un-cracked (Tension) • • • • 38 kd/3 C jd Cracked nAfrp Cross Section f’c = c Ec Transformed Section frp Strain Distribution The concrete is at the elastic stage. Both strain and stress distributions are linear. Triangular stress block with a resultant at it centroid. Lever arm: 𝑗𝑑 = 𝑑 − 𝑘𝑑 Τ3 = 𝑑 1 − 𝑘 Τ3 and T Stress Distribution 𝑗 = 1 − 𝑘 Τ3 Lakehead University 2.3 Flexural Design of FRP-RC Elements Serviceability (Service Stress) b N.A. d Afrp Un-cracked (Compression) c kd/3 kd Un-cracked (Tension) Cracked Transformed Section • Factor (k) for neutral axis depth: 𝑘 = C jd nAfrp Cross Section f’c = c Ec T frp Strain Distribution 2ρ𝑓𝑟𝑝 𝑛𝑓𝑟𝑝 + ρ𝑓𝑟𝑝 𝑛𝑓𝑟𝑝 Stress Distribution 2 − ρ𝑓𝑟𝑝 𝑛𝑓𝑟𝑝 𝑛𝑓𝑟𝑝 = 𝐸𝑓𝑟𝑝 Τ𝐸𝑐 • Service stress in reinforcement is given by: 39 𝑀𝑠 𝑓𝑓𝑟𝑝 = 𝐴𝑓𝑟𝑝 𝑗𝑑 Lakehead University 2.3 Flexural Design of FRP-RC Elements Service Stress Limits Serviceability (Service Stress and Cracking) • Both CSA S806 (CL 7.1.2.2) and CSA S6 (CL 16.8.3) limit the allowable stress in FRP at the service limit state (SLS) by: FSLS × fFRPu Material FSLS CFRP 0.65 AFRP 0.35 GFRP 0.25 Cracking • Cracking in FRP-RC elements is less critical than in steel-RC elements. • Crack width limits are relaxed for FRP-RC elements. 40 Exposure Steel-RC Based on CSA A23.3 FRP-RC From CSA S6 Exterior 0.33 0.50 Interior 0.40 0.70 Lakehead University 2.3 Flexural Design of FRP-RC Elements Limiting Crack Width: Strain Limit Approach Serviceability (Cracking) • To control cracking, both CSA S806 (CL 8.3.1.1) and CSA S6 (CL 16.8.2.3) limit the service strain in the FRP reinforcement to 0.0015. • The service strain can be calculated after calculating the service stress: ε𝑓𝑟𝑝𝑠 = 𝑓𝑓𝑟𝑝𝑠 Τ𝐸𝑓𝑟𝑝 Limiting Crack Width: Crack Control Parameter • When the service strain in the reinforcement exceeds 0.0015, CSA S806 (CL 8.3.1.1) requires that the crack control parameter (z) does not exceed 45,000 N/mm for interior exposure and 38,000 N/mm for exterior exposure. 41 Lakehead University 2.3 Flexural Design of FRP-RC Elements Serviceability (Cracking) Limiting Crack Width: Crack Control Parameter z cracking control parameter, N/mm kb bond dependent coefficient (1.2 for deformed or sand-coated bars) fF stress in the tension FRP reinforcement at location of the crack, MPa dc concrete cover measured from the centroid of the outermost tension reinforcing bar to the extreme tension surface (clear cover, cc ≤ 50 mm), mm A effective tension area of concrete surrounding the flexural tension reinforcement and having the same centroid as that reinforcement, divided by the number of bars, mm2 42 Lakehead University 2.3 Flexural Design of FRP-RC Elements Limiting Crack Width: Crack Width Calculations Serviceability (Cracking) • When the service strain in the reinforcement exceeds 0.0015, CSA S6 (CL 16.8.2.3) requires that the crack width (wcr) does not exceed 0.5 mm for members subjected to aggressive environment and 0.7 mm for other members. wcr fFRP h2 h1 kb dc s 43 crack width at the tensile face of the beam, mm stress in the tension FRP reinforcement at location of the crack, MPa distance from the extreme tension surface to the neutral axis, mm distance from the centroid of outermost tension reinf. to the neutral axis, mm bond dependent coefficient (0.8 for sand-coated bars and 1.0 for deformed bars) concrete cover measured from the centroid of the outermost tension reinforcing bar to the extreme tension surface (clear cover, cc ≤ 50 mm), mm bar spacing, mm Lakehead University Serviceability • Efrp < Esteel Higher deflection with FRP-RCmembers Minimum thickness requirements fromCSAA23.3 are unconservative and not applicable Check FRP-RCmembers against deflection requirements of CSAA23.3 using effective inertia Two approaches to calculate/control deflection: • Minimum thickness of the member • Effective moment of inertia 44 Deflection Serviceability Deflection (1) MinimumThicknessApproach • For steel-RC members, CSAA23.3 (Table 9.2) recommends span (ℓn) to depth (h) ratios Minimumthickness (h) Simply One supported end cont. 45 Both ends cont. Cantilever One-way slabs ℓn /20 ℓn /24 ℓn /28 ℓn /10 Beams ℓn /16 ℓn /18 ℓn /21 ℓn /8 Serviceability MinimumThickness • This value (ℓn / h) can be modified to be applied to an FRP-RC member ℓn h = ℓn s h frp frp Steel d ℓn = member clear span, mm h = member thickness, mm s = maximumstrain allowed in steel at service (0.0012) frp = maximum strain allowed in FRPat service (0.0020) d = coefficient = 0.5 for rectangular section 46 Serviceability: Rectangular Section Deflection (2) Effective InertiaApproach Uncracked (gross) Effective Ig Ie Icr b h3 Ig≈ 12 b(kd)3 Icr= + nfrpAfrp(d-kd)2 3 Ie= k = 47 Cracked It Icr Icr+ 1 – 0.5 Mcr Ma 2 Ig- Icr 2 n f r p + ( n f r p ) 2 − n frp Maximum Permissible Deflections 48 Design Example 2 For the given simply supported beam: – Check the serviceability requirements, including service stress, crack control parameter, and deflection. wDL = 30 kN/m & wLL = 20 kN/m Beamin a building h = 500 mm deff = 455 mm cc = 40 mm (to flexural reinf.) 6-No.10 CFRPbars b = 300 mm 49 Exterior exposure fFRPu = 1,596 MPa f’c = 35 MPa EFRPp = 111 GPa Abar = 71 mm2 dbar = 9.5 mm kb = 1.2 Design Example 2 Step 1: Calculate cracking moment deff = 455 mm h = 500 mm Solution cc = 40 mm 6-No.10 CFRPbars b = 300 mm 3 3 300(500) b h = 3.125 109 mm4 Ig= = 12 12 50 Design Example 2 Step 2: Calculate service moment ws = wDL + wLL ws = 30 + 20 = 70 kN/m Note: No Load Factors At mid-span section: ws 2 50(3) 2 Ms = = = 56.25 kN.m > 44.37 kN.m 8 8 Section is cracked under service load 51 Design Example 2 Step 3: Calculate service stress in CFRPreinforcement k = 2 frp n frp + ( frp n frp )2 − frp n frp n frp = 111, 000 = 4.17 26,622 and frp = and E frp n frp = Ec Afrp 6 71 = = 0.00312 bw d 300 455 k = 2(0.00312)(4.17) + (0.00312 4.17)2 − (0.00312 4.17) = 0.15 52 Design Example 2 Step 3: Calculate service stress in CFRPreinforcement Lever armfactor: j = 1− k 0.15 = 1− = 0.95 3 3 For CFRP: Service limit state is 65%fFRPu = 0.65 × 1,596 = 1,037 MPa> 303.7 MPa OK 53 Design Example 2 Step 4: Control cracking using strain limit approach fFRPs = EFRP FRPs FRPs = fFRPs / EFRP = 303.7 / 111,000 = 0.002736 > 0.0015 We have to calculate crack width 54 Design Example 2 Es f f 3 dc A z = kb Ef deff = 455 mm h = 500 mm Step 5: Calculate crack control parameter cc = 40 mm 6-No.10 CFRPbars kb = 1.2 (CSA S806) b = 300 mm dc = 40 + 9.5/2 = 45 mm 2bd c 2(300)(45) = = 4500 mm2 Since we have one layer, A = n z = 1.2 6 200, 000 (303.7) 3 (45)(4, 500) = 38,100 N/mm 111, 000 z = 38,100 N/mm ≈ 38,000 N/mm OK 55 Design Example 2 Step 6: Calculate cracked moment of inertia h = 500 mm b(kd)3 Icr= + nfrpAfrp(d-kd)2 3 deff = 455 mm cc = 40 mm 6-No.10 CFRPbars b = 300 mm 300(0.15×455)3 Icr= 3 Icr = 0.297109 mm4 56 + 4.17(6×71)(455 - 0.15×455)2 Design Example 2 Step 7: Calculate effective moment of inertia Ie= It Icr Icr+ 1 – 0.5 Mcr Ma 2 It- Icr It Ie= 3.125 0.297 109 0.297 + 1 – 0.5 44.37 56.25 57 ≈ Ig = 3.125 109 mm4 = 0.41 109 mm4 2 3.125 - 0.297 Design Example 2 Step 8: Calculate immediate deflection 5𝑀𝑠 𝑙 2 max = D+L = 48𝐸𝑐 𝐼𝑒 5(56.25 10 6 )(3, 000) 2 = 4.8 mm D+L = 9 48(26, 622)(0.41 10 ) Step 9: Check for permissible deflection δL = δD+L (w L /w D+L) = 4.8 (20/50) = 1.9 mm δL = 1.9 mm< Δallowable = ℓ n / 360 = 3000 / 360 = 8.3 mm 58 Development Length CSA S806 (Clause 9.3.2) k1k 2 k 3 k 4 k 5 = 1.15 d d cs ℓd but dcs ≤ 2.5 db and ff f ' c Ab ≥ 300 mm f’c ≤ 5 MPa Modification factors: (Clause 9.3.3) K1 (bar location factor): 1.3 to 1.0 (300 mm of fresh concrete below ld) K2 (concrete density factor): 1.3 to 1.0 (low to normal) K3 (bar size factor): 0.8 (Ab < 300 mm2), 1.0 (Ab > 300 mm2) K4 (bar fibre factor): 1.0 for GFRP & CFRP, 1.25 for AFRP K5 (bar surface profile factor): 1.0 to 1.8 (sand-coated to indented) 59 Development Length CSA S6 (Clause 16.8.3) & ISIS M3-07 (Sec. 8.3) k k f 1 4 FRP AFRP ≥ 250 mm ℓ𝑙dd = 0.45 d cs + ktr (EFRP / Es ) f cr k4 ( 1.0) is the bar surface factor (ratio of the bond strength of the FRPbar to that of a steel deformed bar having the same Ab). In absence of experimental data, the factor k4 shall be taken as 0.8. (dcs + Ktr Efrp /Es) 2.5db where the transverse reinforcement index Ktr = 0.45Atr fy / (10.5 s n) dcs 60 is the smaller of “concrete cover + db / 2” or “2/3 bar spacing” Development Length Example of calculations: • CFRP bar • No. 16 (diameter 15.9 mm) • f’c = 35 MPa (normal density) • fF = 600 MPa (1.5 times fy steel) • Cover = 35 mm (< 2.5db = 40 mm) • Spacing between bars = 30 mm 61 Development Length Example of calculations: k1k 2 k 3 k 4 k 5 = 1.15 d d cs CSA S806 ff f ' c Ab K1 = 1.0 (location factor) K2 = 1.0 (concrete density) K3 = 0.8 (A = 200 mm2 < 300 mm2) K4 = 1.0 (CFRP) (type of fibre) K5 = 1.0 (sand-coated bar) (surface profile) = 1.15 d 62 (1)(1)(0.8)(1)(1) (600) (200) 35 5 = 630 mm Development Length Development length of bent bars: db lbhf = 165 f c' lbhf = f fu db 3.1 lbhf = 330 63 f c' d ACI 440.1R for ffrpu 520 MPa for 520 < ffrpu < 1040 MPa for ffrpu > 1040 MPa Deformability Steel-RC members < bal crushing 64 > bal yielding crushing no yielding Ductile behaviour Over reinforced Alot of curvature before failure Less deformation before failure Deformability FRP-RC members crushing no yielding FRPs do not yield Alot of curvature before failure Because EFRP < Esteel Important to check deformability of FRPRC members 65 Deformability FRP-RC members = curvature service << ultimate Deformability factor (DF): Curvature and moment at ultimate conditions uM u DF = sM s Curvature and moment at service conditions (εc = 0.001) CSA S6 Approach (Clause 16.8.2.1) & ISIS M3-07 DF ≥ 4 or 6 → Rectangular and T-beams, respectively 66 Deformability FRP-RCmembers • Curvature and moment at service condition, εc = 0.001: fc = c Ec c= 0.001 s = 0.001 kd and where, kd M s = f c b jd 2 u= c cu and jd fc = 0.001 Ec • Curvature and moment at ultimate condition: T frp Strain Distribution a M u = M r = FRP AFRP f FRP d − 2 Where the values of εc & fFRP depend on the mode of failure. 67 C kd /3 kd Stress Distribution Design Example 3 For the given simply supported beam: wDL = 30 kN/m & wLL = 20 kN/m – Calculate the deformability factor and check its adequacy Section information h = 500 mm deff = 465 mm 6-No.10 CFRPbars b = 300 mm 68 Exterior exposure fFRPu = 1,596 f’c = 35 MPa MPa EFRPp = 111 Abar = 71 mm2 GPa kb = 1.2 dbar = 9.5 mm Design Example 3 Solution Step 1: Curvature and moment at service condition: s = 0.001 kd FromExample 3, k = 0.15 d = 465 mm kd M s = f c b jd 2 s = 0.001 = 1.47 10 −5 1.44 0.15 × 465 0.149 455 j = 1− k 0.15 = 1− = 0.95 3 3 (0.15×465) 0.149(455) M s = (0.001 26, 622) 300 (0.95 455) 1066 N.mm 465) = 117 122×10 2 69 Design Example 3 Solution Step 2: Curvature and moment at ultimate condition: u = c cu and FromExample 2, a M u = M r = FRP AFRP f FRP d − 2 cu = a / 1 = 86 / 0.88 → cu = 97.7 mm c = cu = 0.0035 Compression Failure u = 0.0035 = 3.58 10 −5 97.7 and Mr = 193 106 N.mm 70 Design Example 3 Solution Step 3: Deformability factor: 193 × 10 1066 ) u M u (3.58 10 −5 )(190 DF = = M 1.44 × 10-5 122 × 106 3.93 CSAS6 currently requires: DF ≥ 4 for Rectangular-Sections Beam section satisfies deformability requirements 71 Shear Design Cracking Behaviour of Concrete Beams — Along the beam length, the direction of principal tensile stress depends on the magnitude of both normal and shear stresses and ranges between: – Pureshear at 450with the beamaxis (at neutral axis, N.A.) at the proximity of supports, – Pure normal stresses parallel to the beamaxis. 72 Shear Design Mode of Failure Flexural Failure Shear Failure 73 Shear Design Cracking Behaviour of Concrete Beams Diagonal Tension (Shear) Failure 74 Shear Design 75 Shear Design 76 Shear Design 77 Shear Design Types of Shear Reinforcement 78 Vertical Stirrups Bent Bars Inclined Stirrups Welded-Wire Fabric Shear Design Most of the design codes and guides recommend the simplified approach of Vcf + Vsf to shear design Vr = Vcf + Vsf where, Vr = factored shear resistance Vcf = concrete contributionto shear strength Vsf = shear reinforcement contribution to shear strength 79 Shear Design Concrete Contribution Va = Aggregate Interlock ( Interface Shear) C ha V = concrete cz in compression T Vd = Dowel S 80 Force Vc = Vcz + Va + Vd Z Vcz = uncracked concrete shear resistance: ~ 35% Va = aggregate interlock force: ~ 45% Vd = Dowel action of longitudinal reinforcement: ~ 20% Shear Design Shear Reinforcement Contribution Functions of Shear Reinforcement Main Functions: − Carries part of shear (Vs) − Enhances aggregate interlock − Improve longitudinal bar dowel action Secondary Functions: ✓ Dowel action through the inclined crack ✓ Confinement of concrete in compression 81 Shear Design 255 mm 200 = r Plaster bag m m 1 . 38 200 = r Load cell 900 mm m m 0 90 Rollers Elevation 255mm mm 255 Plan FRP stirrup 82 Hydraulic jack Concrete block d = 9.5 d = 9.5 m1 m 8 . 3 Steel plate S806 – Shear Design Provisions Clause 8.4.4 CSAS806 Standard 𝑉𝑟 ≥ 𝑉𝑓 Vr = shear resistance Vf = factored shear force 𝑉𝑟 = 𝑉𝑐 + 𝑉𝑠𝐹 For FRPstirrups 𝑉𝑟 = 𝑉𝑐 + 𝑉𝑠𝑠 For steel stirrups 83 S806 – Shear Design Provisions CSAS806-12 Clause 8.4.4.5 The concrete contribution to shear strength Vc is calculated using the following equations: 1 ' 3 c Vc = 0.05c km kr ( f ) bwd v Where, Such that 0.11 𝜆 𝜙𝑐 84 𝑓𝑐′ 𝑏𝑤 𝑑𝑣 ≤ 𝑉𝑐 ≤ 0.22 𝜆 𝜙𝑐 𝑓𝑐′ 𝑏𝑤 𝑑𝑣 S806 – Shear Design Provisions Clause 8.4.4.5 Where, 𝐸𝐹 is the modulus of elasticity of longitudinal FRPreinforcement 𝑑 is the depth to the centroid of flexural reinforcement 𝑑𝑣 𝑏𝑤 𝑀𝐹 𝑉𝑓 is the effective depth for shear (greater of 0.9 d or 0.72 h) is the width of the web is the ultimate bending moment at section under consideration is the corresponding ultimate shear force at section under consideration 𝜌𝐹𝑤 is the longitudinal reinforcementratio 85 CSAS806-12 S806 – Shear Design Provisions CSAS806-12 Clause 8.4.4.6, 7 &8 Size Effect Factor, ks For members with an effective depth greater than 300 mm and with transverse reinforcement less than the minimum, Vc should be multiplied by the size effect factor, ks whichis given as: 750 ks = 1.0 450 + d 86 S806 – Shear Design Provisions • Shear carried by transverse reinforcement CSAS806-12 Clause8.4.4.9 For FRPstirrups For steel stirrups where, 𝑨𝒗 or 𝑨𝑭𝒗 =Area of shear reinforcement, =Ultimate strength of FRPshear reinforcement. It shouldnot be 𝒇𝑭𝒖 taken greater than0.005 Ef 𝒇𝒚 =Yield strength of steel reinforcement, s θ 87 = Spacing of shear reinforcement. =Angle of the compressionstrut S806 – Shear Design Provisions CSAS806-12 Clause8.4.4.9 The angle θ of the compressive stress shall be calculated as follows: 300 ≤ θ = 30 + 7000 ε𝓵 ≤ 600 where the longitudinal strain (εl) at the mid-depth of the section shall be calculated as: Mf = dv + (V f −V p ) + 0.5N f − Ap f po 2(E F AF + E p Ap ) Mf +V f For non-prestressed members dv = without axial load: 2(EF AF ) 88 S806 – Shear Design Provisions CSAS806-12 Clause8.4.5 Minimumshear reinforcement (Vf > 0.5 Vc) bw s Av ,min 0.07 f 0.4 f fu ' c Maximumshear reinforcement spacing Clause 8.4.6 The maximum spacing between the transverse reinforcement shall not exceed 0.6 dv cot θ or 400 mm 89 S806 – Shear Design Provisions CSAS806-12 Clause8.4.4.4 Maximumshear strength The nominal shear resistance Vr shall not exceed: Vr 0.22c f c'bw d v + 0.5V p + (M dcV f ) / M f Where, 𝑀𝑑𝑐 is the decompression moment, equal to the momentwhenthe compressive stress onthe tensile face of aprestressed member is zero For non-prestressed members, 90 Vr 0.22c f c' bwd v S806 - Design Example Example: d = 460 mm fc‘ = 35 MPa (Normal weight concrete) Ec = 28.1 GPa wDL = 40 kN/m wLL = 30 kN/m Longitudinal GFRPbars - Bar diameter, db = 15.9 mm - Afl (longitudinal) = 792 mm2 - ffu = 1,000 MPa - Ef = 60 GPa h = 540 mm Transverse GFRPstirrups - 91 Bar diameter, db = 12.7 mm - Ab (transverse) = 127 mm2 - ffu = 600 MPa - Ef = 50 GPa bw = 280 mm S806 - Design Example Factored Load: wu = 1.25 wDL + 1.5 wLL wu = 1.25 (40 kN/m) + 1.5 (30 kN/m) wu = 95 kN/m Shear and moment at critical section: (at a distancedv awayfrom the support) d v = 0.9d = 0.9(460) = 414 mm or d v = 0.72h = 0.72(540) = 390 mm Vf = wf L/ 2- wf dv Vf = 95 x 3.8 /2 - 95 x 0.414 = 141.2 kN Mf = wf L.dv / 2- wf dv2/2 Mf = 95 x 3.8 x 0.414 /2 – 95 x 0.4142/ 2 = 66.6 kN.m 92 S806 - Design Example Determination of Vc : 1 ' 3 c Vc = 0.05c k m k r ( f ) bwd v Fw = Af bd = 792 = 0.006 280 460 1 1 k r = 1+ (E f Fw )3 = 1+ (60,000 0.006) 3 = 8.11 Vf d 141.2 0.46 = km = = 0.99 < 1.0 ok M 66.6 1 3 Vc = 0.05(1)(0.65)(0.99)(8.11)(35) (280)(414) 10−3 = 98.6 kN 93 S806 - Design Example Check limits for Vc : 0.11 𝜆 𝜙𝑐 𝑓𝑐′ 𝑏𝑤 𝑑𝑣 ≤ 𝑉𝑐 ≤ 0.22 𝜆 𝜙𝑐 𝑓𝑐′ 𝑏𝑤 𝑑𝑣 0.11 𝜆 𝜙𝑐 𝑓𝑐′ 𝑏𝑤 𝑑𝑣 = 0.11(1)(0.65) 35(280)(414)×10-3 = 49 kN 0.22 𝜆 𝜙𝑐 𝑓𝑐′ 𝑏𝑤 𝑑𝑣 = 0.22(1)(0.65) 35(280)(414)×10-3 = 98 kN Since 49 kN < 𝑉𝑐 = 98.6 kN but > 98 kN 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡𝑎𝑘𝑒 𝑉𝑐 = 98 kN Since Vf = 141.2 kN > Vc = 98 kN, Shear reinforcement is required 94 S806 - Design Example Determine VsF : Determine stirrup spacing, s, required for Vsf : and 300 ≤ θ = 30 + 7000 εl ≤ 600 6 Mf 66.6 10 +V f + 141.2 10 3 dv 414 = l = = 0.0032 0.003 2(E F AF ) 2(60,000 792) 95 S806 - Design Example Determine stirrup spacing, s, required for Vsf : θ = 30 + 7000 εl = 30 + 7000 (0.003) θ = 510 300 ≤ θ ≤ 600 OK Considering 2-branch stirrups, 𝒇𝑭𝒖 ≤ 0.005 Ef = 𝟎. 𝟎𝟎𝟓 (𝟓𝟎, 𝟎𝟎𝟎) = 250 MPa 96 S806 - Design Example Check other requirements for spacing, s 𝑏𝑤𝑠 𝐴 𝑣,𝑚𝑖𝑛 = 0.07 𝑓𝑐′ 0.4 𝑓𝐹𝑢 2 × 127 ≥ 0.07 280 𝑠 35 0.4 (250) smax ≤ 220 mm s ≤ 0.6 dv cot θ = 0.6 (414) cot 51 = 229 mm s ≤ 400 mm Use s = 150 mm 97 S806 - Design Example Check maximum shear strength: Vr = Vc +VsF 0.22c f c'bwd v For the selected stirrup spacing, s = 150 mm VsF = 43.2 kN Vr = 98.0 + 43.2 = 141.2 kN 0.22(0.65)(35)(280)(414) 10−3 = 580 kN OK 98 S806 - Design Example Determine shear reinforcement zones: The shear strength corresponding to smax = 220 mmis given by: Vr ,min = Vc +VsF ,min = 98.0 + 29.4 = 127.4 kN starting at distance x = [ (95 x 1.9) - 127.4] / 95 = 0.54 m 99 fromsupports S806 - Design Example Final design: Use single-looped No.13 (12.7-mm diameter) GFRP stirrups: - spaced at 150 mm (from support to 600 mm each side); - spaced at 220 mm (in the middle 2600 mm). No.13 @ 150 mm 600 mm 100 No.13 @ 220 mm 2600 mm No.13 @ 150 mm 600 mm Dr. Ahmed Bediwy, PEng Assistant Professor, Dept. of Civil Engineering Office: F1-1087 101 Lakehead University
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