CIVL 228 Introduction to Structural Engineering Spring 2015 38Normal

# Civl 228 introduction to structural engineering

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CIVL 228 Introduction to Structural Engineering, Spring 2015 38 Normal distribution A statistical model to describe the probability of a distribution -4 -2 0 2 4 6 8 10 0 0.05 0.1 0.15 0.2 0.25 0.3 X m s s s s Area under m ± 2 s = 95% Area under m ± s = 68%
CIVL 228 Introduction to Structural Engineering, Spring 2015 39 Safety indexA successful design will have membercapacitiesthatexceed the memberdemandsthroughout the life of thestructure. But how do we ensure that the capacityexceeds the demand considering the variability in both thecapacity and demand described above?
CIVL 228 Introduction to Structural Engineering, Spring 2015 40 Demand vs. CapacityNormalized probability density function (pdf):-1000100 200 300400 500600 7008000123456x 10-3DemandCapacityHow do we calculate the probability of failure?
CIVL 228 Introduction to Structural Engineering, Spring 2015 41 Safety index pdf for the failure: Lets assume both Capacity, C, and Demand, D, are normal distribution. Hence its probability can be modeled using the standard “bell” curve. Let X = C-D Hence, when X<0 D>C the system failed. Using basic probability theory μ X = μ C - μ D • σ X = sqrt ( σ C 2 + σ D 2 ) • β X = μ X / σ X
CIVL 228 Introduction to Structural Engineering, Spring 2015 42 Safety index P(system failure) = P(X): -200 -100 0 100 200 300 400 500 600 700 0 0.5 1 1.5 2 2.5 3 3.5 x 10 -3 Capacity - Demand [kN], X pdf [-] Probability of failure [ m X = 246, s X = 126] β X σ X , where β X = the safety index. X = 0 X = μ X Area under the curve = probability of failure.
CIVL 228 Introduction to Structural Engineering, Spring 2015 43 What is the safety index,bX?
CIVL 228 Introduction to Structural Engineering, Spring 2015 44 Safety index Load and resistant factors: • β X σ X = μ X μ X = μ C - μ D • σ X = sqrt ( σ C 2 + σ D 2 ) β X ( σ C 2 + σ D 2 )= μ C - μ D β X σ C 2 C = - μ D - β X σ D 2 μ C (1 - β X σ C 2 C ) = μ D (1 + β X σ D 2 D ) To have a safe design: f should be < 1 and g should be > 1. Hint: σ /μ = COV = coefficient of variance. f = resistant factor g = load factor
CIVL 228 Introduction to Structural Engineering, Spring 2015 45 Example of load factors Load combinations (NBCC 2005): Case Load Combination (2,3) Principal Loads Companion Loads (9) 1 1.4D 2 (1.25D (7) or 0.9D (4) ) + 1.5L (5) 0.5S (1) or 0.4W 3 (1.25D (7) or 0.9D (4) ) + 1.5S 0.5L (1,6) or 0.4W 4 (1.25D (7) or 0.9D (4) ) + 1.4W 0.5L (6) or 0.5S 5 1.0D (4) + 1.0E (8) 0.5L (1,6) + 0.25S (1) Check NBCC 2005 Section 4.1.3.2 for details.
CIVL 228 Introduction to Structural Engineering, Spring 2015 46 Example of resistant factors Concrete (ACI): Tension: 0.9 Compression: 0.75 Bending: 0.9 Shear: 0.85 Steel (AISC): Tension yielding: 0.9 Tension fracture: 0.75 Compression: 0.85 Bending: 0.85 Shear: 0.9 Why different f for different loading cases? Lower f higher b . More confident in certain failure mode.
CIVL 228 Introduction to Structural Engineering, Spring 2015 47 Structural failure (snow load) A factory building in North America
CIVL 228 Introduction to Structural Engineering, Spring 2015 48 Structural failure (wind load) Hurricane Katrina, New Orleans (2005) Levee fails
CIVL 228 Introduction to Structural Engineering, Spring 2015 49 Structural failure (Hurricane wind load) Hurricane Katrina, New Orleans (2005)

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