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e. What is the required nominal strength of the column for a safety factor of 1.67?

e. What is the required nominal strength of the column for a safety factor of 1.67?

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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A column in the upper story of a building is subjected to a compressive load from the following sources: Dead Load= 308kN Occupancy Live Load= 170kN Roof Live Load= 187kN Wind Load= 197kN The answer from A to D is the attached pictures. Kindly answer the letter E. thankyou a. If load and resistance factor design it to be used, determine the factored load (required strength) to be used in the design of the column. Which NSCP load combination controls? b. What is the required design strength of the column? c. What is the required nominal strength of the column for a resistance factor of 0.90? d. If allowable strength design used, determine the required load capacity (required strength) to be used in the design. Which NSCP load combination controls? e. What is the required nominal strength of the column for a safety factor of 1.67?
According to NSCP, following factored load combinations are valid: 1) P = 1.4D = 1.4(308) = 431.2 kN 2) P = 1.2D + 1.6L + 0.5 L₁ = 1.2(308) + 1.6(170) + 0.5(187) 735.1 kN = 3) P = 1.2D + 1.6L₁ + L = 1.2(308) + 1.6(187) + 170 = 838.8 kN (controls) 4) P = 1.2D + 1.6L + 0.5W = 1.2(308) + 1.6(187) + 0.5(197) 767.3 kN 5) P = 1.2D + L + W + 0.5 L₁ = 1.2(308) + 170 + 197 + 0.5(187) = 830.1 kN 6) P = 0.9D + W = 0.9(308) + 197 = 474.2 kN From above calculation, the largest value will be the required design strength of column. Therefore, Pu = 838.8 KN
Transcribed Image Text:According to NSCP, following factored load combinations are valid: 1) P = 1.4D = 1.4(308) = 431.2 kN 2) P = 1.2D + 1.6L + 0.5 L₁ = 1.2(308) + 1.6(170) + 0.5(187) 735.1 kN = 3) P = 1.2D + 1.6L₁ + L = 1.2(308) + 1.6(187) + 170 = 838.8 kN (controls) 4) P = 1.2D + 1.6L + 0.5W = 1.2(308) + 1.6(187) + 0.5(197) 767.3 kN 5) P = 1.2D + L + W + 0.5 L₁ = 1.2(308) + 170 + 197 + 0.5(187) = 830.1 kN 6) P = 0.9D + W = 0.9(308) + 197 = 474.2 kN From above calculation, the largest value will be the required design strength of column. Therefore, Pu = 838.8 KN
d) We know, Pu=$Pn Here, LRFD column resistance factor = 0.90 => Pn = Pu $ 838.8 0.9 = 932 KN
Transcribed Image Text:d) We know, Pu=$Pn Here, LRFD column resistance factor = 0.90 => Pn = Pu $ 838.8 0.9 = 932 KN
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