Titel_SS06

The mean value and standard deviation of the safety margin M are thus:

M350 200 150 MPa

352240 53.15 MPa
M 

whereby the reliability index may be calculated as:

150

2.84

53.15



Finally the failure probability is determined as:

( 2.84) 2.4 10^3

PF

,    

Example 6.2 – Error Accumulation Law

As an example of the use of the error propagation law consider a right angle triangle ABC,
where B is the right angle. The lengths of the opposite side b and adjacent side a are
measured. Due to measurement uncertainty the length of the sides a and b are modelled as
independent Normal distributed random variables with expected values a = 12.2, b = 5.1

and standard deviations a = 0.4 and b = 0.3, respectively. It is assumed that a critical

condition will occur if the hypotenuse c is larger than 13.5 and the probability that this
condition should happen is to be assessed.

Based on the probabilistic model of a and b the statistical characteristics of the hypotenuse
c given by:

cab^22

may be assessed through the error propagation model given by Equations (6.13)-(6.14),
yielding:





0

22

2

22

1 22 22

()

i

ab

n

Xa

i i

Ec

Var c fa
x ab ab



2

b

b

 



!

 "

"#$% xx

x^

which by inserting for and a b their expected values yields:





22

22

22 22

12.2 5.1 13.22

12.2 5.1

0.182

12.2 5.1 12.2 5.1

ab

Ec

Var c 

 





As seen from the above the variance of the hypotenuse c depends on the chosen linearization
point. If instead of the mean value point a value corresponding to the mean value plus two
standard deviations was chosen the variance of would have been: c

 222222

(^13) 0.4 5.7 0.3 0.149
13 5.7 13 5.7
Var c