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# Lec8 MECH ENG STRucture

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### Lec8 MECH ENG STRucture

1. 1. 2.001 - MECHANICS AND MATERIALS I Lecture #8 10/4/2006 Prof. Carol LivermoreRecall from last lecture:Find: u(x), FA , FB , FC1. Equilibrium Fu = 0 P − FA − F B − F C = 0 MA = 0 Fc L Pa − − FB L = 0 22. Force-Deformation FA = kδA 1
2. 2. FB = kδB FC = kδC3. Compatibility δA = uA y δB = uA + L tan ϕA y z L δC = uA + y tan ϕA z 2New this lecture: Small Angle Assumption: A sin ϕz ϕA A tan ϕz = A ≈ z = ϕA z cos ϕz 1For small ϕA : Arc length ≈ a straight line displacemtn in y. zRewrite compatibility. δA = uA y δB = uA + LϕA y z A L A δA = uy + ϕ 2 zSubstitute compatibility into force-deformation A FA = kuy FB = k(uA + LϕA ) y z 2
3. 3. L A FC = k(uA + y ϕ ) 2 zSubstitute this result into equilibrium equations: L A P − kuA − k(uA + LϕA ) − k(uA + y y z y ϕ )=0 2 z L L Pa − k(uA + ϕA ) −L k(uA + LϕA ) = 0 y 2 2 z y zSolve: 3 P = 3kuA + Lkϕz y A 2 3 5 P a = LkuA + L2 kϕA y z 2 4Divide by k. P 3 = 3uA + LϕA y k 2 zDivide by Lk/2. 2P a 5 = 3uA + LϕA y Lk 2 zSo: −P 2a ϕA = z 1− Lk LSubstitute: P P 2a uA = y + 1− 3k 2k L P P 2a P 2a u(x) = + (1 − ) − (1 − )x 3k 2k L Lk LUNIAXIAL LOADING Behavior of a uniaxially loaded bar
4. 4. L = LD + δ P = kδ is a property of the bar. FORCE STRESS = = σ( Like Pressure) UNIT AREA NUnits: 1 m2 = 1 Pa (SI Units) lbsUnits: 1 in2 = PSI (English Units) P σ= AEngineering Stress For small deformation P σ= A0 CHANGE IN LENGTHSTRAIN = LENGTH = δ = LStrain is dimensionless.Engineering Strain (For small deformations) δ = L0 4
5. 5. Stress-Strain plotFor Uniaxial Loading: σ=EMaterial property is E, Young’s Modulus.Note: Units of E = Pa, E 109 Pa (or GPa).So, what is k for uniaxiail loading? σ=E P σ= A0 δ = L0 So: P Eδ = A0 L0 EA0 P = δ L0 So: EA0 k= for uniaxial loading L0Deformation and Displacement
6. 6. δ (du(x) + dx) − dx) = = L dx du(x) = dx u(x) = axial displacement of x du(x) (x) = dx σ(x) (x) = ESo: du(x) σ(x) P = = dx E AE δ L P du = dx 0 0 AE δ L Px u = 0 AE 0 PL δ= AESo: AEδ P = LWhat are some typical values for E? E Steel 200 GPa Aluminum 70 GPa Polycarbonate 2.3 GPa Titanium 150 GPa Fiber-reinforced Composites 120 GPaSelection of material? Optimize k for a particular A Steel: ks = k = As Es L Al: kA = k = AA EA LSame k ⇒ As Es = AA EA Es So: AA = As EA So: AA ≈ 3As ⇒ the aluminum is three times bigger 6
7. 7. May need to optimize weight (think about airplanes) ⇒ need to include den-sity.What happens if you keep pulling on a material? σy = Yield Stressp = Plastic Strain - Not Recoverable.e = Elastic Strain - Fully Recoverable.t = e + p = Total StrainWhat about pulling on a bar in uniaxial tension? 7