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Ch 2 - Part 3

# Ch 2 - Part 3 - Chapter 2 RC Design Evaluation and Design...

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1 Chapter 2 – RC Design Evaluation and Design RC Rectangular Beam Problems BRAE/ITRC Cal Poly Part 3 Strength Design Factored Loads U = 1.2D + 1.6L 1.4D strength furnished strength required BRAE/ITRC Cal Poly U 1.2D + 1.6L Note: D and L are “service” loads

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2 Strength Reduction Factor Φ 0 9 b di = 0.9 bending Φ = 0.75 shear BRAE/ITRC Cal Poly Practical moment strength = φ Mn Strength Reduction Factor Φ = special case, ε y < 0.005 0.900 ε y < 0.005 ε y 0.700 0.750 0.800 0.850 Beta Φ BRAE/ITRC Cal Poly 0.600 0.650 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 fc' (psi)
3 Beam Analysis 14,000 lbs Given 2-3ss: f’c = 4,000 psi, A615 grade 60 steel wLL = 800 lb/ft Plus wDL = 800 lb/ft BRAE/ITRC Cal Poly Steps to Solve Ultimate Moment, M n 1. N C = N T N c = 0.85 fc’ a b, N T = A s fy T 2. 0.85 fc’ a b = A s fy 3. solve for a = A s x f y / (0.85 x f’ c x b) 4. Z = d – a/2 5. solve for M n = N C (d – a/2) BRAE/ITRC Cal Poly

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4 Rectangular Beam Evaluation write φ M n equation in terms of b, d, and As given A s , b and d ( ρ =A s /bd) φ M n = φ bd 2 k /12 (k in Tables A-7 through A-11) solve for φ M solve for n 1) Givens: b = 12 in, d = 17.5 in, A s = 4.0 in 2 , f’ c = 4,000 psi, fy = 60,000 psi 2) Get ρ = A s /bd and Get k : k = 0.949 ksi ( = 949 psi) 3) Check φ based on ε t where φ = 0.65 + ( ε t - 0.002) x (250/3) BRAE/ITRC Cal Poly 4) Solve φ M n = φ bd 2 k /12 = 0.8675 (12in) (17.5in) 2 * 949psi /(12in/ft) = 252,000 ft-lbs 5) ck φ M n against M u : φ M n > M u =w u l 2 /8 + P u l/4 Practical Moment Strength From 2 3ss e ample problem From 2-3ss example problem: φ M n
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Ch 2 - Part 3 - Chapter 2 RC Design Evaluation and Design...

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