Introduction to strain gauges and beam bending

Labs and work

• You need to let us know (while working, not after the fact) if labs are taking too long.

• You should be working, but it should be reasonable (3 credit course, ~9hrs/week total).

• Waiting to the night before, means you are up all night.

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 Mpa – How much can ½ x 1/16 inch bar hold in tension?

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

+ -

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

Demo

Left side only provides 2.5 V We already have that available

2.5 (V ref)

10

120 + dR

115

Vmeas

5 V

115

1 2 0

Vmeas

10

5 V

120

120+dR (strain gauge)

=Strain gauge

Strain gauge

A quantitative test

2.5 (V ref)

10

120

115

Vmeas

5 V

100 K

Conclusion – part 1

• Bridge circuit + instrumentation amp. allow us to detect small changes in resistance.

• Mechanical strain proportional to ΔR/R.

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

(1/radius) curvature theis

This relation is easy to prove by geometry

Neutral axis

Cantilever beam

L

P

Isolate the beam – Find the reaction forces

PR Sum forces

L

P

R

Also need a bending moment

PLM

PR

Sum forces

L

P

R

M

Sum moments

State of stress inside the beam

L

P

P

P L

State of stress inside the beam

L

P

P

P L

Imagine a cut in the beam

State of stress inside the beam

PP

P L

x

)( xLPPxPLM

M

Calculate shear and bending moment to hold at equilibrium

Shear and bending moment diagram

X

X

Shear

MP(L- x)

P

PP

P L

x

ML

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

Strain in cantilever – at support

L

P

P

PL

3

121

2)(

bhE

hPL

EI

My

h

b