ExamIISolution

# ExamIISolution - Florida International University...

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Florida International University Reinforced Concrete Design (CES 4720) Fall 2009 Exam II 1. a) The capacity of concrete: (5 points) 𝑉𝑉 𝑐𝑐 = 2 𝑓𝑓 𝑐𝑐 𝑏𝑏 𝑤𝑤 𝑑𝑑 = 29 𝑘𝑘 𝑑𝑑 = 21 3 = 18 ′′ 𝑏𝑏 𝑤𝑤 = 2 × 6 = 12 ′′ b) The capacity of steel: (10 points) 𝑉𝑉 𝑠𝑠 = 𝐴𝐴 𝑣𝑣 𝑠𝑠 � 𝑓𝑓 𝑦𝑦 𝑑𝑑 For the first 4’ 𝑉𝑉 𝑠𝑠 = 4 × 0.11 4.5 × 60,000 × 18 1000 = 105.6 𝑘𝑘 For the next 3’ 𝑉𝑉 𝑠𝑠 = 4 × 0.11 9 × 60,000 × 18 1000 = 52.8 𝑘𝑘 c) The capacity of the beam: For the first 4’ 𝑉𝑉 𝑢𝑢 ≤ 𝜙𝜙𝑉𝑉 𝑛𝑛 = 𝜙𝜙𝑉𝑉 𝑐𝑐 + 𝜙𝜙𝑉𝑉 𝑠𝑠 = 101 𝑘𝑘 (15 points) For the next 3’ 𝑉𝑉 𝑢𝑢 ≤ 𝜙𝜙𝑉𝑉 𝑛𝑛 = 𝜙𝜙𝑉𝑉 𝑐𝑐 + 𝜙𝜙𝑉𝑉 𝑠𝑠 = 61.4 𝑘𝑘 For the last 3’ 𝑉𝑉 𝑢𝑢 𝜙𝜙𝑉𝑉 𝑐𝑐 2 = 10.9 𝑘𝑘

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d) Check 𝑨𝑨 𝒗𝒗 𝒔𝒔 𝒎𝒎𝒎𝒎𝒎𝒎 For the first 4’ 𝐴𝐴 𝑣𝑣
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