By Victor E Saouma
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Extra info for Machanics and design of reinforced concrets
5 in, determine the reinforcement As and possibly As . Solution: 1 In this problem, unfortunately an iterative method diverges if we were to start with a = d5 . 8 Moment-Curvature Relations 2–33 1. 111-g) Thus compression steel is required. 2. 112-d) 3. 113-d) Note that if it turned out that fs < fy , then we will need to make an assumption on As (such as As = A2s , as we will have three equations (2 of equilibrium and one of strain compatibility) and four unknowns (As , As , fs and c). 8 Moment-Curvature Relations In ordinary reinforced concrete design, we need not be concerned by the moment curvature relation of a ﬂexural member.
55) 7. 59) 8. 60) 9. 65) 10. 10) Equivalent width for uniform stress, Fig. 2): 38 1. 1 2 (b − bw ) ≤ 8hf 2. b < 4bw for isolated T beams only 3. hf > 4. b < 39 bw 2 L 4 Two possibilities: 1. Neutral axis within the ﬂanges (c < hf ) ⇒ rectangular section of width b, Fig. 9. 2. Neutral axis in the web (c > hf ) ⇒ T beam. 9: T Beam as Rectangular Section For T beams, we have a large concrete area, start by assuming that failure will occur by steel yielding, Fig. 10. 40 41 The approach consists in decomposing As into 2 components Fig.
Check from tables that the selected beam width will provide room for the bars chosen with adequate cover and spacing. 1. 2. 3. 4. 5. Victor Saouma Mechanics and Design of Reinforced Concrete Draft 2–14 Bar Size #3 #4 #5 #6 #7 #8 #9 #10 #11 #14 #18 FLEXURE Nominal Diam. 6 in single layer of reinf. 39-e) Note: Victor Saouma Mechanics and Design of Reinforced Concrete Draft 2–16 FLEXURE 1. 87 in 2. 7 = 144 3. 3 in. 45) εmax where fc corresponds to εmax . 1. 003. (a) Determine the equation for the exact stress distribution on the section.