Introduction to strain gauges and beam bending

Beam bending

Galileo, 1638 (though he wasn’t right)

Normal stress (σ) and strain (ε)

P P

L

A

P

L

δ

Stress-strain

Yield stress in “ordinary” steel, 400 Mpa How much can 1 x 1cm bar hold in tension?Yield stress in “ordinary” Aluminum 100

Hooke’s law

steelfor GPa 200E

modulus sYoung'or modulus elastic is

E

E

What is the strain just before steel yields?

Strain gauge

6.4x4.3 mm

Gauge factor

R is nominal resistanceGF is gauge factor. For ours, GF = 2.1

Need a circuit to measure a small change in resistance

Wheatstone bridge

R4

R 1

Vmeas

VCC

R2

R3

+ - 4

Our setup

120

1 2 0

Vmeas

5 V

120

120+dR (strain gauge)

Strain gauge

Proportional to strain !

In practice we need variable R.Why?

115

1 2 0

Vmeas

10

5 V

120

120+dR (strain gauge)

Strain gauge

Beams in bending

Beam in pure bending

MM

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

MM

“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

y=0

Beam in pure bending

Fig 5-7, page 304

Beam in pure bending

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

Neutral axis

Normal stress in bending

M

Take a slice through the beam

σ

Neutral axis is the centroid

y

Flexure formulaWill derive this in Mechanics of Solids and Structures

EI

MyI

My

b

h

12

3bhI

CrossSection

y