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LRFD-Steel Design

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LRFD-Steel Design

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    1. LRFD-Steel Design Dr. Ali I. Tayeh second Semester

    2. Steel Design Dr. Ali I. Tayeh Chapter 2

    3. Design philosophies Structural member design means selection of a cross section that will be more safe and economic to resist the applied force Economic usually means minimum weight There are three different philosophies to design any structure: Allowable stress design. Plastic design LRFD. Allowable stress design (elastic design working stress design) The member which selected should has cross sectional properties (area, moment of inertia) large enough to prevent the maximum stress from exceed the allowable

    4. Design philosophies Working stress: Stress resulting from working load (service load). For any Designed member will stressed to no more than allowable stress when subjected to working load. Plastic Design: Based on failure consideration conditions Member is selected by using the criterion that mean structure will fail at a load higher than the working load. The total load will be less than the failure loads by a factor of safety called load Factor.

    5. Design philosophies Design procedure: Multiply the working load service load by the load factor to obtain the failure loads. Determine cross sectional proprieties to resist failure under these loads Select the lightest cross sectional properties that has the mentioned properties.

    6. Design Philosophies LRFD (Load and Resistance Factor Design): Failure condition is considered Load factor are applied to service load Resistance factor reduce the theoretical strength Factored load = factored strength Load load Factor = Resistance Resistance factor load Factor = 1 Resistance factor = 1

    7. Load and Resistance Factors: Factored load = factored strength S ?i Qi = Rn Where Each effect have a different load factor Load factor Value for a particular load effect will depend on combination of load under consideration. The load factors and load combinations are based on extensive statistical studies.

    8. Load and Resistance Factors: Load combinations: 1.4(D+F). 1.2(D+F+T)+1.6(L+H)+ 0.5(Lror S or R). 1.2D+1.6(Lror S or R)+(0.5L or 0.8W). 1.2D+1.6W+0.5L +0.5(Lror S or R). 1.2D+1.0E+0.5L+0.25. 0.9D+1.6W+1.6H. 0.9D+1.0E+1.6H.

    9. Load and Resistance Factors: Load combinations: Where:

    10. Load and Resistance Factors: Normally, fluid pressure F , earth pressure H, and self strain force T are not applicable to the design of structure steel members, so load combination become as: 1.4(D). 1.2(D) + 1.6(L)+ 0.5(Lror S or R). 1.2D + 1.6(Lror S or R)+(0.5L or 0.8W). 1.2D + 1.6W+0.5L +0.5(Lror S or R). 1.2D 1.0E+0.5L+0.25. 0.9D (1.6W or 1.0E).

    11. Load and Resistance Factors: Example:2.1 Dead load ( D ) = 109 kips compression. Floor live load ( L ) =46 kips compression. Roof live load ( Lr ) =19 kips compression. Snow ( S ) = 20 kips compression. Find the controlling load combination and the corresponding factor load What is the required Nominal strength ? If the resistance factor , equal 0.85

    12. Load and Resistance Factors: Example:2.1 (Cont.). the controlling load combination is that procedure largest factored load from the following load combination Combination 1 1.4(D). = 1.4 109=152.5 kips. Combination 2 1.2(D) + 1.6(L)+ 0.5(Lror S or R). Because S larger than Lr we will use S instead of Lr 1.2(D) + 1.6(L)+ 0.5(S) =1.2 109 +1.6 20 + 0.5 46 =214.4 kips

    13. Load and Resistance Factors: Example:2.1 (Cont.). Combination 3 1.2D + 1.6(Lror S or R)+(0.5L or 0.8W). 1.2D + 1.6(S )+(0.5L) = 1.2 109+ 1.6 20+0.5 46 = 185.5 kips Combination 4 1.2D + 1.6W+0.5L +0.5(Lror S or R) 1.2D + 0.5L +0.5( S ) =1.2 109 + 0.5 46 +0.5 20 Smaller than combination 3

    14. Load and Resistance Factors: Example:2.1 (Cont.). Combination 5 1.2D 1.0E+0.5L+0.25. Smaller than combination 4 Combination 6 0.9D (1.6W or 1.0E). Note that dead load was multiplied to 0.9 this was reduce it , so this case is smaller than any combination. Answer Combination 2 controls, and factored load = 214.4 kips

    15. Load and Resistance Factors: Example:2.1 (Cont.). S ? Q = Rn S ? Q=214 kips , =0.85 ?Rn =(214 0.85 =252.2kips) Rn = 252.2kips Answer.

    16. Manual of Steel Construction Part 1 , part 2, part 3 , part 4 , .. Part 17 See Text Book.

    17. Introduction -Steel Design Home work: Page 33 . Text book

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