Introduction to strain gauges and
beam bending
Beam bending
Galileo, 1638 (though he wasn’t right)
Normal stress (σ) and strain (ε)
L
δ
P
P
P

A


L
Stress-strain
Yield stress in “ordinary” steel, 400 Mpa
Yield stress in “ordinary” Aluminum 100
How much can 1 x 1cm bar hold in tension?
Hooke’s law
  E
E is elastic modulus or Young's modulus
E  200 GP a for steel
What is the strain just before steel yields?
Strain gauge
6.4x4.3 mm
Gauge factor
R is nominal resistance
GF is gauge factor. For ours, GF = 2.1
Need a circuit to measure a small change in resistance
Wheatstone bridge
VCC
R3
R1
+
R2
-
4
V
meas
R4
Our setup
5V
120
Strain gauge
120+dR (strain gauge)
V
meas
120
120
Proportional to strain !
In practice we need variable R.
Why?
5V
Strain gauge
120
120+dR (strain gauge)
V
meas
120
115
10
Beams in bending
Beam in pure bending
M
M
DaVinci-1493
"Of bending of the springs: If a straight spring is bent,
it is necessary that its convex part become thinner
and its concave part, thicker. This modification is
pyramidal, and consequently, there will never be a
change in the middle of the spring. You shall discover,
if you consider all of the aforementioned
modifications, that by taking part 'ab' in the middle of
its length and then bending the spring in a way that
the two parallel lines, 'a' and 'b' touch a the bottom,
the distance between the parallel lines has grown as
much at the top as it has diminished at the bottom.
Therefore, the center of its height has become much
like a balance for the sides. And the ends of those
lines draw as close at the bottom as much as they
draw away at the top. From this you will understand
why the center of the height of the parallels never
increases in 'ab' nor diminishes in the bent spring at
'co.'
Beam in pure bending
M
M
y=0
“If a straight spring is bent, it is necessary that
its convex part become thinner and its concave
part, thicker. This modification is pyramidal,
and consequently, there will never be a change
in the middle of the spring.” DaVinci 1493
Beam in pure bending
Fig 5-7, page 304
Beam in pure bending
Neutral axis
Lines, mn and pq remain straight –
due to symmetry.
Top is compressed, bottom expanded,
somewhere in between the length is
unchanged!
 
y

This relation is easy to prove by geometry
Normal stress in bending
M
σ
y
Take a slice through the beam
Neutral axis is the centroid
Flexure formula
Will derive this in Mechanics of Solids and Structures
My

I
My

EI
y
Cross
Section
h
b
3
bh
I
12