Lec9 MECH ENG STRucture

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Lec9 MECH ENG STRucture

  1. 1. 2.001 - MECHANICS AND MATERIALS I Lecture #9 10/11/2006 Prof. Carol LivermoreReview of Uniaxial Loading:δ = Change in length (Positive for extension; also called tension)Stress (σ) σ = P/A0Strain ( ) at a point L − L0 δ = = L0 L0Stress-Strain Relationship ⇒ Material Behavior σ=E 1
  2. 2. Force-Displacement Relationship P = kδFor a uniaxial force in a bar: EA0 k= L0Deformation and Displacement Recall from Lab: The springs deform. The bar is displaced.EXAMPLE: 2
  3. 3. du(x) (x) = dx (x)dx = duSign ConventionTrusses that deform Bars pinned at the joints How do bars deform? How do joints displace?EXAMPLE:Q: Forces in bars? How much does each bar deform? How much doespoint B displace?Unconstrained degrees of freedom 1. uB x 2. uB y 3
  4. 4. Unknowns 1. FAB 2. FBCThis is statically determinate (Forces can be found using equilibrium)FBD: Fx = 0 at Pin B −FAB − FBC sin θ = 0 Fy = 0 −P −P − FBC cos θ = 0 ⇒ FBC = cos θ P −FAB + sin θ = 0 cos θSo: FAB = P tan θForce-Deformation Relationship P = kδ EA k= L FAB δAB = kAB FBC δBC = kBC 4
  5. 5. AE kAB = L sin θ AE kBC = LSo: P tan θ δAB = AE L sin θ −P cos θ δBC = AE L P L sin θ tan θ δAB = AE −P L δBC = AE cos θCheck:Compatibility 5
  6. 6. Algorithm to find B 1. If we only have uB , what δ AB and δ BC would result? x 2. If we only have uB , what δ AB and δ BC would result? y 3. What is the total δ AB and δ BC is I have both uB and uy ? x B 4. Solvefor uB and uB from known δ AB and δ BC . x yStep 1 δ BC = uB sin θ x LN EW = BC (L + δ1 )2 + Δ2 BC 2 2δ1 δ BC Δ2 = L (1 + + 1 2 )+ 2 L L L √ x 1+x≈1+ L B BC 2 δ1 C δ1 Δ2 LN EW =L 1+ + + L 2L2 2L2 2 BC δ1 →0 2L2 Δ2 →0 2L2For δ << L: δ BC ≈ DSo: BC Lnew = L + δ1 6

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