Report

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 modulusor Young's modulus E 200GPa 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