Why need probabilistic approach? Rain probability
How does that affect our behaviour?? ?
Summary of Common Results (Cont’d)
if X1 and X2 are N(m1, s1) and N(m2, s2) respectively
Z is N(mz, sz)
where
if X1 and X2 are s.i. r12 = 0 if Xi = N(mi, si) ; i = 1 to n
Z is N(mz, sz)
where
E4.8
S = D + L + W Column with capacity, R mD = 4.2 , sD = 0.3 , dD = 7%
mL = 6.5 , sL = 0.8 , dL = 12%
mW = 3.4 , sW = 0.7 , dW = 21% S = total load
= D + L + W mS = mD + mL + mW = 14.1
sS = 0.32 + 0.82 + 0.72 =1.1
a) P(S > 18) = 1 – F
= 1 – F(3.55) = 0.000193 Assume D, L, W are s.i.
E4.8 (Cont’d)
P(failure) = P(R < S)
= P(R – S < 0)
= P(X < 0) R = N(mR, dRmR)
where mR = 1.5 mS
= 1.5 x 14.1
= 21.15
dR = 0.15
R = N(21.15, 3.17) mX = -14.1 + 21.15 = 7.05
sX = 1.12 + 3.172 = 3.36
P(F) = F = F(-2.1) = 0.018 X = R – S mR – design capacity
1.5 – design safety factor, SF
Recall:
mX = mR - 14.1
sX = (0.15mR)2 + 1.12
= F-1(0.001) = – 3.09
0.0225mR2+1.21 = 20.8 – 2.95mR + 0.105mR2
0.0825 mR2 – 2.95 mR +19.59 = 0
mR = 8.812 or 26.9
Set: P(F) = F = 0.001 Since mR should be larger than 21.5
mR = 26.9
Check mR = 8.812
P(F) = F
= F(3.09) 1.0
If the target is P(F) = 0.001 mR = ? and assume dR = 0.15
or 1.
Total weight = T = 2W
Normal with
mT = 2mW = 200
sT = 2sW = 40 Total weight = T = W1 + W2
Normal with
mT = 200
sT =
=
=
if s.i. and s1 = s2 = 20 2. ? T W = weight of a truck = N(100, 20)
We are interested in the total weight of 2 trucks
E4.10
Sand
Footing
P
S Sand Property – M Footing Property – B and I Assume P, B, I, and M are s.i. and log-normal with parameters lP, lB, lI, lM and zP, zB, zI, zM, respectively
Central Limit Theorem
S will approach a normal distribution regardless the individual probability distribution of Xi if N is large enough
Taylor Series Approximation
First order approximation:
g(mX) g’(mX) (X – mX) X g(X) mX X mx - mx
Var(X) …
Observe validity of linear approx depends on:
1) Function g is almost linear, i.e. small curvature
2) sx is small, i.e. distribution of X is narrow
Uses
1. Easy calculations
2. Compare relative contributions of uncertainties – allocation of resource
3. Combine individual contributions of uncertainties
Reliability Computation
Case 1: If R, S are normal
where mZ = mS – mR and sZ = sS2 + sR2 Case 2: If R, S are lognormal
where lZ = lS – lR and zZ = zS2 + zR2 Suppose R denotes resistance or capacity
S denotes load or demand
Satisfactory Performance = {S < R}
PS = P(S < R) and Pf = 1 - PS
Case 3: If R is discrete, S is continuous
Example on Case 3
S = N(5, 1)
r P(R = r)
5
6
7 0.1 0.3 0.6
Reliability – Based Design
If b PS Reliability
b = Reliability index
= F-1(PS) Design mR = mS + bsS2 + sR2 b Observe for Case 1 in which R and S are both Normal
Example:
S = N(5, 2)
R = N(mR, 1)
mR = ?
Require Pf = 0.001 or PS = 0.999
b = F-1(0.999) = 3.1
Design mR = 5 + 3.122 + 12 = 11.93
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