lec22 - 2.001 - MECHANICS AND MATERIALS I Lecture #24...

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2.001 - MECHANICS AND MATERIALS I Lecture 12/4/2006 Prof. Carol Livermore Torsion (Twisting) M xx = M t Examples where torsion is important: - Screwdriver - Drills -Prope l lers Use superposition Example: Unifrom along length, has circular symmetry 1 #24
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ϕ ( x ) = Angle of Twist (Total) at Point x (Built Up) Compatibility: d = ϕ ( x + dx ) R ϕ ( x ) R d = γdx So: = ϕ ( x + dx ) R ϕ ( x ) R = dϕR γ = R dx What is γ ? So: γ = γ Shear Strain Thus: γ = R dx 2
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± ± ± And: γR d ϕ ± θx == 22 dx ± rr = ± θθ = ± xx = ± = ± rx =0 Constitutive Relations: σ θx =2 θx = = Gr , where G is the shear modulus . dx E G = 2(1 + ν ) σ xx = σ θθ = σ rr = σ = σ rx σ θx = Gr dx Ey Recall beam bending σ xx = ρ . Equilibrium: ² M x M t + rdF A dF = σ θx dA M t + θx dA A M t = rGr dA dx A 3
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± ± ± ± ± M t = Gr 2 dA dx A Recall beam bending: 1 M = Ey 2 dA ρ A Note: :wha thappen s ² dx Gr 2 dA : efective stifness A M t : what we apply IF G is constant ”Special Case” M t = Gr 2 dA dx A J r 2 dA Polar Moment oF Inertia A So For ”special case” G is constant. M t = GJ dx Recall beam bending: EI M = ρ IF G is not constant: ( GJ ) eff = Gr 2 dA A When G is constant: M t = GJ and generally σ θx = Gr dx dx M t σ θx = GJ Gr So: M t r σ θx = J 4
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±² Example for J 1. Cicular solid shaft ³³ 2
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lec22 - 2.001 - MECHANICS AND MATERIALS I Lecture #24...

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