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Mechanics 1L

Mechanics 1L - 2.001 MECHANICS AND MATERIALS I Lecture#9...

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± == 2.001 - MECHANICS AND MATERIALS I Lecture #9 10/11/2006 Prof. Carol Livermore Review of Uniaxial Loading: δ = Change in length (Positive for extension; also called tension) Stress ( σ ) σ = P/A 0 Strain ( ± )atapo int L L 0 δ L 0 L 0 Stress-Strain Relationship Material Behavior σ = 1

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Force-Displacement Relationship P = For a uniaxial force in a bar: EA 0 k = L 0 Deformation and Displacement Recall from Lab: The springs deform. The bar is displaced. EXAMPLE: 2
±± du ( x ) ± ( x )= dx ± ( x ) dx = du Sign Convention Trusses 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 does point B displace? Unconstrained degrees of freedom 1. u 2. u B x B y 3

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± ± Unknowns 1. F AB 2. F BC This is statically determinate (Forces can be found using equilibrium) FBD: F x =0atPinB F AB F sin θ =0 F y P P F cos θ F = cos θ P F AB +s i n θ cos θ So: F AB = P tan θ Force-Deformation Relationship P = EA k = L F AB δ AB = k AB F δ =

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Mechanics 1L - 2.001 MECHANICS AND MATERIALS I Lecture#9...

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