Mos 10-2 strain

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Transcript of Mos 10-2 strain

  • 1. Introduction to strain gauges andbeam bending

2. Beam bendingGalileo, 1638 (though he wasnt right) 3. Normal stress () and strain ()P PPA=e dLs=L 4. Stress-strainYield stress in ordinary steel, 400 Mpa How much can 1 x 1cm bar hold in tension?Yield stress in ordinary Aluminum 100 5. Hookes laws =Eeis elastic modulus or Young's modulus=EE 200 GPa for steelWhat is the strain just before steel yields? 6. Strain gauge6.4x4.3 mm 7. Gauge factorR is nominal resistanceGF is gauge factor. For ours, GF = 2.1Need a circuit to measure a small change in resistance 8. Wheatstone bridgeR 4R 1VmV C Ce a sR 2R 3+ - 4 9. Our setupStrain gauge1 2 01 2 0Vm5 Ve a s1 2 01 2 0 + d R ( s t r a i n g a u g e )Proportional to strain ! 10. In practice we need variable R.Why?Strain gauge1 1 51 2 0Vme a s1 05 V1 2 01 2 0 + d R ( s t r a i n g a u g e ) 11. Beams in bending 12. Beam in pure bendingM M 13. DaVinci-1493"Of bending of the springs: If a straight spring is bent,it is necessary that its convex part become thinnerand its concave part, thicker. This modification ispyramidal, and consequently, there will never be achange in the middle of the spring. You shall discover,if you consider all of the aforementionedmodifications, that by taking part 'ab' in the middle ofits length and then bending the spring in a way thatthe two parallel lines, 'a' and 'b' touch a the bottom,the distance between the parallel lines has grown asmuch at the top as it has diminished at the bottom.Therefore, the center of its height has become muchlike a balance for the sides. And the ends of thoselines draw as close at the bottom as much as theydraw away at the top. From this you will understandwhy the center of the height of the parallels neverincreases in 'ab' nor diminishes in the bent spring at'co.' 14. Beam in pure bendingM MIf a straight spring is bent, it is necessary thatits convex part become thinner and its concavepart, thicker. This modification is pyramidal,and consequently, there will never be a changein the middle of the spring. DaVinci 1493y=0 15. Beam in pure bendingFig 5-7, page 304 16. Beam in pure bendingLines, mn and pq remain straight due to symmetry.Top is compressed, bottom expanded,somewhere in between the length isunchanged!e = - yrThis relation is easy to prove by geometryNeutral axis 17. Normal stress in bendingMTake a slice through the beamNeutral axis is the centroidy 18. Flexure formulaWill derive this in Mechanics of Solids and StructuresMyIMyEI=s=ebhbh3 I =12CrossSectiony