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Exam III solution

# Exam III solution - Florida International University...

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Florida International University Reinforced Concrete Design (CES 4720) Fall 2009 Exam III 1. (5 points) 𝜌𝜌 g = A s A g = 1% < 8% 1% 𝑂𝑂𝑂𝑂 5 points 2. (55 points) 𝜀𝜀 𝑠𝑠 2 = 𝜀𝜀 𝑦𝑦 = 𝑓𝑓 𝑦𝑦 𝐸𝐸 𝑠𝑠 = 0.00207 5 points 𝜀𝜀 𝑠𝑠 2 = 𝑑𝑑 2 −𝑐𝑐 𝑐𝑐 𝜀𝜀 𝑐𝑐𝑐𝑐 = 0.00207 10 points 𝑐𝑐 = 10.34 ′′ 5 points 𝜀𝜀 𝑠𝑠 1 = 𝑐𝑐−𝑑𝑑 1 𝑐𝑐 𝜀𝜀 𝑐𝑐𝑐𝑐 = 0.00228 > 𝜀𝜀 𝑦𝑦 𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑑𝑑 5 points 𝛽𝛽 1 = 0.825 5 points 𝑎𝑎 = 𝛽𝛽 1 𝑐𝑐 = 8.55 ′′ 5 points 𝑃𝑃 𝑛𝑛 = 0.85 𝑓𝑓 𝑐𝑐 𝑏𝑏𝑎𝑎 + 𝐴𝐴 𝑠𝑠 1 �𝑓𝑓 𝑦𝑦 0.85 𝑓𝑓 𝑐𝑐 � − 𝐴𝐴 𝑠𝑠 2 𝑓𝑓 𝑦𝑦 = 646 𝑘𝑘 10 points 𝑀𝑀 𝑛𝑛 = 0.85 𝑓𝑓 𝑐𝑐 𝑏𝑏𝑎𝑎 � 2 𝑎𝑎 2 + 𝐴𝐴 𝑠𝑠 1 �𝑓𝑓 𝑦𝑦 0.85 𝑓𝑓 𝑐𝑐 � � 2 − 𝑑𝑑 1 � − 𝐴𝐴 𝑠𝑠 2 𝑓𝑓 𝑦𝑦 2 − 𝑑𝑑 2 = 457 𝑘𝑘 − 𝑓𝑓𝑓𝑓 10 points 𝑌𝑌 = 𝑀𝑀 𝑛𝑛 𝑃𝑃 𝑛𝑛 = 8.5

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