Introduction to solid mechanics

baconossifiedMechanics

Oct 29, 2013 (3 years and 7 months ago)

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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

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

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

+

-

4

Our setup

Strain gauge

Proportional to strain !

In practice we need variable R.

Why?

Strain gauge

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

“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!

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 formula

Will derive this in Mechanics of Solids and Structures

b

h

Cross

Section

y