The following loads were computed for designing a column in a frame: • Dead Load = 200 K • Roof Live Load = 50 K • Live Load from floors = 250 K • Compression Wind = 128 K, Tensile Wind = 104 K • Compression Earthquake = 60 K, Tensile Earthquake = 70 K Determine the critical design column load PU using LRFD combinations.
The following loads were computed for designing a column in a frame: • Dead Load = 200 K • Roof Live Load = 50 K • Live Load from floors = 250 K • Compression Wind = 128 K, Tensile Wind = 104 K • Compression Earthquake = 60 K, Tensile Earthquake = 70 K Determine the critical design column load PU using LRFD combinations.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
The following loads were computed for designing a column in a
frame:
• Dead Load = 200 K
• Roof Live Load = 50 K
• Live Load from floors = 250 K
• Compression Wind = 128 K, Tensile Wind = 104 K
• Compression Earthquake = 60 K, Tensile Earthquake = 70 K
Determine the critical design column load PU using LRFD combinations.
Expert Solution
Step 1: Introduction and given data
Given values,
- Dead load, D = 200 k
- Live load from floors, L = 250 k
- Load from roof, Lr=50 k
- Compression wind = 128 k
- Tension wind = 104 k
- compression earthquake=60k
- tensile earthquake=70k
- Snow load = 0
- Rain load = 0
D = dead loads
E = earthquake loads
L = live loads (floor)
Lr = live loads (roof)
R = rain load
S = snow load
W = wind load
To determine the critical design column load PU using LRFD combinations
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Follow-up Question
The critical design column load, Pu is 665 k How did you find the Pu
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