Errors in EI Measurement due to Shaft Weight
Dave
Tutelman  August 28, 2011
Instruments
that measure EI tend to use substantial forces to load the
shaft,
forces way more than
the weight of the shaft. For that reason, the shaft weight is usually
ignored, on the assumption that it is negligible compared to the
loading force. Recently, Frans van Daalen brought to my
attention
a
research paper from 1998 that found substantial measurement errors near
the tip and butt due to the weight of the overhanging shaft section.
This article finds an estimate for such errors.
Here are the major results, for those who don't want to wade through
the math:
 The error due to
overhang for typical shafts may be as much
as 8%. That is not a seriously large number, but is higher than most
sources of error in EI measurement.
 The error due to weight
in the measured span is
negligible,
typically in the vicinity of 0.1%.
 Such
errors do not occur if you use differential deflection to do your
measurements. That is, using both a preload and
a
fullload measurement,
and subtracting one from another, will eliminate this source of error,
and many others as well. Note that Russ Ryden's EI instrument, as well
as mine and Don
Johnson's, use
differential deflection. If you use one of these designs, then shaft
weight errors are of academic interest only. Why it
happens is
explained later in this article.
Analysis
Figure
1. Idealized view of EI machine
Figure 1 shows the basic assumption for measuring EI, in this case,
near the tip of the shaft. The shaft is supported at either side of a
span of length S, and a measurement force F is applied at the middle of
the span. The deflection due to the force is measured, and used to
calculate EI. The calculation for deflection y
is done using the wellknown formula:
This is the idealized assumption for measuring EI. Note that the weight
of the shaft is never considered. A more complete view of the
measurement would consider shaft weight, and would look like Figure 2.
Figure
2. More complete view includes shaft weight
Even this model is slightly idealized, assuming a
uniform weight distribution over the length of the shaft. For most
shafts, this is close to accurate. True, the wall thickness changes
over the length of the shaft. But it tends to be thickest where the
shaft has the largest diameter; the resulting weight distribution is
much
more uniform than you might imagine.
Considering shaft weight adds two sources of error:
 Overhang
Error  The weight outside the span contributes a moment
that will raise the point where deflection is mesured, reducing the
measured deflection.
 Span
Error  The weight inside the span contributes a load that
will lower the point where deflection is measured, adding to the
measured deflection.
The main work of this article is to estimate overhang and span errors.
For our sample calculations, we choose
values that tend to emphasize the error. A longer, heavier shaft gives
more overhang error, and somewhat more span error. For the parameters
for our computation, we will use:
Parameter 
Value 
Comment 
L 
48
inches 
Longer
shaft gives more overhang error. 
W 
80
grams 
Heavier
shaft gives more overhang error.
Also slightly more span error. 
S 
10
inches 
Typical
for EI machine.
See article for pros
and cons. 
F 
20
pounds
= 9100 grams 
This
is a "middling" value for known EI machines.
Higher values give less error, but risks deforming
the shaft and introducing other forms of error. 
Note
that we are using a mixed system of units (inches and grams). It is
worth worrying about, but in the end does not hurt us. Reasoning:
 We
are consistent with our units for length (inches) and force (grams).
Note that grams are force (not mass) in all our computations.
 Our final result is
percentage error, which is dimensionless. It gets that way because
every length is cancelled out (numerator and denominator), as is every
force. So any conversion factor we needed to "fix" the units would also
be canceled out. So the
percentage error in this case is not affected by the mixed units.
Overhang Error
Figure
3. Overhang error
Figure 3 shows the factors involved in computing overhang error. We
will compute the deflection y_{overhang}
due to the overhanging weight. In this section, we will only consider
the overhanging weight, nothing else. Then we will compare that
deflection to our ideal deflection y_{ideal}.
The result is the overhang error percentage.
The
overhang weight contributes a moment  and only a moment  to the
deflection in the middle of the span. Because we do not include any
load in the span here, the
bending moment is constant across the whole span.
The trick to computing this
moment is to recognize that:
 The force acts through the middle of the overhang, so the
moment arm is ½(LS).
 The amount of the force is the weight of the overhang, the
proportion of the shaft weight not over the span: W(LS)/L.
So the moment M
is
M
= ½ (LS) 
W
(LS)
L 
=

W(LS)^{2}
2
L 
The
shaft across the span is deflected by a constant moment, implying no
shear at all. That is not a configuration in the usual handbooks, so we
will have to derive it. We derive the deflection following the approach
in Timoshenko's classic text. (S.
Timoshenko, "Strength
of Materials", Van Nostrand 1940, Chapter V) We start from
the basic principle:
Since M
is constant over the span, this should be simpler than Timoshenko's
example. We start by multiplying both sides by dx
and integrating:
where
C is the constant of integration. We evaluate C
by knowing that
the maximum deflection, and therefore zero slope, is at the middle of
the span. So
we set the slope dy/dx
to zero at x=S/2.
0 =
MS/2 + C
C =
MS/2
Thus:
EI 
dy
dx 
=
Mx + MS/2 = M (S/2  x) 
So now we know the slope of the shaft between the spans. To get the
deflection itself, we again multiply by dx
and integrate.
EI
y =
M ( 
Sx
2 
 
x^{2}
2 
)
+ C_{1} 
Once again, we need to evaluate the constant of integration. We are
looking for maximum deflection at x=S/2,
and the deflection is zero at x=0
and x=S.
Let's set y=0
at x=0.
(Note that we ought to get the same value for C_{1} if
we had set y=0
at x=S,
and that is exactly what happens.)
0 = 0
+ C_{1
}C_{1} = 0
Plugging this back into our last integration result:
EI
y =
M ( 
Sx
2 
 
x^{2}
2 
) 
We want the deflection in the middle of the span, so set x=S/2
and solve for y.
y
= 
M
EI 
( 
S^{2}
4 
 
S^{2}
8 
)
= 
MS^{2}
8 EI 
Now all we have to do is plug in the formula we already computed for M.
y
= 
S^{2}
8 EI 
W(LS)^{2}
2
L 
=

W
S^{2}(LS)^{2}
16
L EI 
We're
all set to compute the percentage error. That should be the overhang
deflection computed here, divided by the ideal deflection computed
earlier.
y_{overhang}
y_{ideal} 
= 
W
S^{2}(LS)^{2}
16
L EI 
* 
48
EI
FS^{3} 
y_{overhang}
y_{ideal} 
=

3W
F 

(LS)^{2}
LS^{
} 
Plugging in the values from our original table, we find that y_{overhang} / y_{ideal} is
0.079.
So the overhang error is 7.9%. Not an overwhelming amount, but still
one of the largest sources
of error in EI measurement.
(To assure myself that I did it right, I derived it another way and got
the same answer. For those who are interested, the alternative derivation is in
the Appendix.)
Span Error
Figure
4. Span error
Figure 4 shows the model for computing the span error. This turns out
to be very easy, because there is a wellknown formula for it.
The density ρ
is already known to be W/L,
so:
As before, we will compute the percentage error. That should be the
span
deflection computed here, divided by the ideal deflection computed
earlier.
y_{span}
y_{ideal} 
= 
5
W S^{4}
384
L EI 
* 
48
EI
FS^{3} 
y_{span}
y_{ideal} 
=

5
W
8
F 

S
L^{
} 
Plugging in the values from our original table, we find that y_{span} / y_{ideal} is
0.00114.
So the overhang error is about a tenth of a percent. Negligible!
Why
Differential Deflection Avoids Overhang Error
The statement is made above that differential deflection avoids
overhang error completely (and span error, but that's negligible
anyway). How does it do that?
Shaft bend is additive. If you separate out the bends due to each
source and compute them separately, you can get the total bend by just
adding them together. Engineers call this the Principle of Superposition, and
use it all the time to compute elastic deformation. (In fact, we
implicitly used it in the calculations above; now we will use it
explicitly.) Superposition says that we can separately compute y_{ideal},
y_{overhang},
and y_{span},
and just add them together (signed addition, of course) to get the
total
deflection Y.
When we use differential deflection, we compare the actual measured deflection
under two different conditions: with a preload and with a full load.
Call these deflections Y_{pre}
and Y_{full}.
So, in differential deflection, we actually measure twice.
Y_{pre}
= y_{ideal}(preload) + y_{overhang}
+ y_{span}
Y_{full}
= y_{ideal}(full load) + y_{overhang}
+ y_{span}
Then we subtract Y_{pre} from
Y_{full} to
get our working deflection. For our working load, we subtract the
preload from the full load. Notice what happens:
 The deflection y_{overhang}
is
the same for both preload and full load, because it depends only on
the weight of the shaft, not how much you load the shaft.
 Similarly for the deflection y_{span},
and for the same reason.
 So the subtraction causes y_{overhang}
to
cancel out, and likewise y_{span},
leaving only y_{ideal}(full
load  preload).
So differential deflection measurements are immune to overhang and span
error.
Appendix
Alternate
derivation of overhang error
Timoshenko's treatment of bending
starts with the observation that the bend at any point along the shaft
is a curvature of some radius. The radius of curvature at that point on
the shaft is
given by EI/M
(where M
is, of course, the bending moment at that point on the shaft). But we
are in the very fortunate position that overhang error results from a
constant bending moment across the span S.
Therefore, the bend is simply a circular arc of radius r.
So let's deal with it as a geometrytrigonometry
problem, rather than the calculus solution above.
The figure to the
right shows the geometry problem we are solving:
 The red arc is the shaft between the span supports (the
black circles).
 The dashed line is a straight line across span S.
 There are three radii shown, from the center of curvature
to the arc. Each is of the same length r because the bending
moment is constant.
 The middle radius is divided into two segments, x
and y,
by the span line (dashed).
 Similarly, the span line is bisected into two halves, each
of
length S/2.
 Since the curvature is greatly exaggerated in this diagram
(you'd never bend a shaft anywhere near this much to measure its EI),
the angles a
are in fact much smaller than shown. (This is important because we will
be using smallangle approximations in the derivation.)
The deflection we want to find is y
in the diagram. So let's start with:
y = r
 x
We know that r
is related pretty simply to the bending moment, but finding x
is a geometry problem. We solve it by observing the trigonometry
relations:
Now we are going to use the fact that a is a very small angle, so we
can approximate sine and cosine as:
sin
x ≅ x and
cos
x ≅ 1  x^{2}/2
Plugging these back into our equations for sine and cosine above, we
get:
1
 
a^{2}
2 
=
1  
S^{2}
2
* 4 r^{2} 
=

x
r 
We can solve the last of these for x,
which is what we are looking for.
1
 
S^{2}
2
* 4 r^{2} 
=

x
r 
Now we just apply this to our original proposition: y = r  x
y
= r  x =
r  r  
S^{2}
8r 
Notice, it has just been a geometry problem so far, no physics. Now we
use Timoshenko's bending curvature radius (r = EI/M),
and plug it into the last equation:
We already figured out the M
due to overhanging weight.
Plugging this into our equation for y:
y
= 
W
S^{2} (LS)^{2}
16
L EI 
This is exactly the same as the equation for y
that we got in the section on overhang error, so that answer is now
more trusted.
