56
Chapter 2
EXAMPLE 2.12. Examine Example 2.11 using a card sort to obtain
the standard deviation Jr. In Example 2.11
kt r = 6667 ohms
and
~ = 300 ohms with ~2 = 600
in order to obtain RTmin multiply ~1 and ~2 by 2.576 standard deviations and
subtract from #1 and/~2
R1 minR2 rnin
(9227)(18,454)
RTmi, = Rlmi n q_ Rzmi n -- (9227) + (18,454) = 6151 ohms.
From Eq. (2.26) and from Fig. 2.4 and Table 2.2 for two card sorts
X= 4.056.
#T -- R~’m~n = XZc
6667 - 6151
~c --- 127.22 ohms
4.056
The percentage error compared to the exact solution for ~c
149.1 - 127.22
percentage error =
149.1
× 100 = 14.68% on the low side
C. Computer Estimate of Variance and Distribution
When an equation has several variables and the distributions of each of
these variables can be determined, it is possible to use the distribution
of the variables to computer generate and graphically determine what form
the equation takes. This would also give an independent check on validity
for Eqs. (2.16), (2.25), and (2.26) where no previous experience is available.
Further, the computer generated data could be used in a solution sizing
parts in a coupling equation Eq. (2.42).
V. SAFETY FACTORS AND PROBABILITY OF FAILURE
The applied load f(a) is held in equilibrium by a resisting capacity f(A) of
which both will have a distribution due to the variables not being considered
as constant values. The desired condition is that the capacity is always
greater than the load and the overlap coupling of the two distributions Fig.
2.6 is a small failure value. These should be prescribed values set by the
design criterion. The failure values can be found by computer analysis
for distributions other than Gaussian or normal functions. However, when
Chapter 2
EXAMPLE 2.12. Examine Example 2.11 using a card sort to obtain
the standard deviation Jr. In Example 2.11
kt r = 6667 ohms
and
~ = 300 ohms with ~2 = 600
in order to obtain RTmin multiply ~1 and ~2 by 2.576 standard deviations and
subtract from #1 and/~2
R1 minR2 rnin
(9227)(18,454)
RTmi, = Rlmi n q_ Rzmi n -- (9227) + (18,454) = 6151 ohms.
From Eq. (2.26) and from Fig. 2.4 and Table 2.2 for two card sorts
X= 4.056.
#T -- R~’m~n = XZc
6667 - 6151
~c --- 127.22 ohms
4.056
The percentage error compared to the exact solution for ~c
149.1 - 127.22
percentage error =
149.1
× 100 = 14.68% on the low side
C. Computer Estimate of Variance and Distribution
When an equation has several variables and the distributions of each of
these variables can be determined, it is possible to use the distribution
of the variables to computer generate and graphically determine what form
the equation takes. This would also give an independent check on validity
for Eqs. (2.16), (2.25), and (2.26) where no previous experience is available.
Further, the computer generated data could be used in a solution sizing
parts in a coupling equation Eq. (2.42).
V. SAFETY FACTORS AND PROBABILITY OF FAILURE
The applied load f(a) is held in equilibrium by a resisting capacity f(A) of
which both will have a distribution due to the variables not being considered
as constant values. The desired condition is that the capacity is always
greater than the load and the overlap coupling of the two distributions Fig.
2.6 is a small failure value. These should be prescribed values set by the
design criterion. The failure values can be found by computer analysis
for distributions other than Gaussian or normal functions. However, when