This is a sample uncertainty calculation for the MEL 2 lab by Matt Young. There are
3 parts: Uncertainty of yield strength, uncertainty of modulus of elasticity,
systematic error due to compliance of machine.
Version 1, March 13, 2002.
See also Overheads on uncertainty analysis
Copyright © 2002 by Matt Young. All rights reserved.
Matt Young’s Home Page
1. Yield strength.
Suppose that your graph look something like this:
S tress (a rb itra ry u nits )
E stim a te o f Y
P e ak -to -pe a k n o ise ,
or 2 ∆Y
N
S lo p e = E (fitte d to
lin ear part o f cu rve
on ly)
S tre ss = 0
-3
S tra in (x 1 0 )
You will of course have many more points. We estimate the yield strength by eye and
get, say, 10 000 lb/5 in2, or Y = 2 ksi.
(a) Component of uncertainty due to uncertainty of area A:
Y ∝A
∆Y A
∆A
=
= 0 .04 (from u n certain ty ov erh ead s)
Y
A
∆A
∆Y A = Y ×
= 0 .0 4 Y
A
u A = 0 .0 4 Y / 3 = 0 .0 4 × 2 k si / 3 ≅ 5 0 p si
This is the first component of uncertainty.
(b) Component of uncertainty due to noise:
We bracket the horizontal part of the curve as shown and estimate the confidence
interval 2Y as (say) 20 psi (noise is greatly exaggerated in the figure). Thus,
∆Y N = 20 psi / 2 = 10 psi, so
u N = 10 psi / 3 ≅ 6 psi
This is the second component of uncertainty.
(c) Component of uncertainty due to force:
Force is proportional to pressure in hydraulic system. Specification of transducer is
0.25 % (does not include your calibration). Let’s guess that the overall
uncertainty of the calibration is 1 %. Then,
∆Y P
= 0 .0 1
Y
∆Y P = 0 .0 1 × Y = 0 .0 1 × 2 ksi = 2 0 psi, so
u P = 2 0 psi / 3 ≅ 1 2 psi
This is the third component of uncertainty.
[It would be better to calculate the slope uncertainty of the calibration curve and use
that value as the uncertainty in place of 1 %. You would not divide that value by
3 (why not?).]
(d) Suppose that these are all the errors we can think of. The combined standard
uncertainty is
u c = u A 2 + u N 2 + u P 2 = 5 0 2 + 6 2 + 1 2 2 p si ≅ 5 0 p si
The expanded uncertainty is 2 × uc, so
Y = 2 ksi ± 10 0 psi, or
Y = 2 00 0 ± 10 0 psi
This is a relative uncertainty of about 5 %. The numbers are made up. But if your own
calculations of uncertainty give a few percent relative uncertainty, you may have
to explain a larger discrepancy between your values and handbook values.
I strongly recommend that you set up your uncertainty calculations in MathCad or
equivalent, since you will be able to use the same program several times.
2. Modulus of elasticity.
We calculate E by fitting a line of best fit (trend line) to the linear part of the data, being
careful not to fit to too long a range. Suppose that E = 10 Msi.
(a) Component of uncertainty of E due to stress ):
E ∝ σ and σ ∝ A , so
∆E σ ∆A
=
= 0 .0 4 , as ab o ve, so
E
A
u σ = 0 .0 4 × 1 0 M si / 3 ≅ 2 3 0 k si
This is the first component of uncertainty.
(b) Component of uncertainty due to strain :
Suppose that the strain is given by = /d, where is elongation and d = 4 in is the
length of the specimen. Suppose further that the maximum elongation is 0.2 in,
so the maximum strain is 0.2 in/4 in = 0.05 in. Let us guess that we can measure
length within 0.003 in. Then = 0.003 in and
E ∝1 / ε , so
∆E ε ∆ε 0 .0 0 3 in
= 0 .0 1 5 , o r
=
=
ε
E
0 .2 in
∆E ε = 0 .0 1 5 × E = 0 .0 1 5 × 1 0 M si = 1 5 0 k si
u ε = 1 5 0 k si / 3 ≅ 9 0 k si
This is the second component of uncertainty.
(c) Component of uncertainty due to slope:
Use Taylor’s formulas (see overheads) to calculate the standard deviation )B of the
slope. Suppose that )B = 200 ksi. Then,
u B = σ B = 2 0 0 k si ( n o d ivisio n b y
3)
This is the third component of uncertainty.
(d) Suppose that these are all the errors we can think of. The combined standard
uncertainty is
u c = u σ 2 + u ε 2 + u B 2 = 2 30 2 + 9 0 2 + 2 00 2 k si ≅ 3 20 k si
The expanded uncertainty is 2 × uc, so
Y = 10 M si ± 640 ksi, or
Y = 10 000 ± 640 ksi
Note that our estimate of 90 ksi due to strain error would have to be several times
higher to be significant, so we are justified not refining that calculation.
3. Systematic error due to compliance of machine.
Givens: Length of specimen L = 6 in. Diameter D = 1/2 in. Modulus of elasticity E =
10 Msi (guess). Overall length of rods holding specimen L1= 15 in. Diameter of
rods holding specimen D1 = 1.5 in. Modulus of elasiticty E1 = 10 Msi (guess).
If the force is 5000 lb, then the elongation of the specimen (assuming that it is still in
the elastic region) is
A = π × ( 1 / 2 in ) 2 / 4 ≅ 0 .2 in 2
σ = 5 0 0 0 lb / 0 .2 in 2 ≅ 2 5 0 0 0 p si = 2 5 k si
δ = L × σ / E = 6 in × 2 5 k si / 1 0 M si = 0 .0 1 5 in
But the elongation of the rods is
A ′ = π × ( 1.5 in ) 2 / 4 ≅ 1.8 in 2
σ ′ = 5 0 0 0 lb / 1 .8 in 2 ≅ 2 8 0 0 p si = 2 .8 k si
δ ′ = L ′ × σ ′ / E ′ = 1 5 in × 2 .8 k si / 1 0 M si = 0 .0 0 4 in
Net elongation of specimen = measured elongation – elongation of rods. Elongation of
rods is a systematic error. In this example (with made-up numbers), the relative
systematic error is about 25 %. It shows one reason that measurement of E
using the displacement transducer is inaccurate.
Systematic error has uncertainty due to unknown value of E, which may have a
confidence interval of (say) ±10 % and which contributes to the uncertainty
budget.
A thousand apologies to my former colleagues at NIST for making them use SI units
exclusively.
4. Cookbook.
Shhh. Don’t tell. I’m supposed to make you struggle.
(a) If you can calculate a standard deviation, calculate it. That is, if you have a
statistical sample, calculate the standard deviation of the mean ). For a
stationary distribution, that means the ordinary standard deviation; for a line, it
means the standard deviation of the slope.
(b) All other errors: Estimate or guesstimate the largest possible error. (That value
defines the half-width of the confidence interval.) Divide it by
3 . The result is
the component of uncertainty. Calculate all the relevant components of
uncertainty u1, u2, u3, ... .
(c) If there are any significant corrections (systematic errors), make them. Calculate
the uncertainty of the correction as in (a) and (b). Usually it is negligible unless
the correction is very large.
(d) Calculate the combined standard uncertainty
(e) Express the result as M ± 2 u c .
.
u c = σ 2 + u 12 + u 22 + u 32 +