EGM 3520 Chapter 10 Columns Contents 10.1 Introduction 10.2...

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EGM 3520 Chapter 10 Columns
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EGM 3520 Department of Mechanical and Aerospace Engineering Contents 10.1 Introduction 10.2 Stability of Structures 10.3 Euler’s Formula for Pin- Ended Columns 10.4 Extension of Euler’s Formula to Columns with Other End Conditions 10.5 Eccentric Loading ; the Secant Formula 10.6 Design of Columns under a Centric Load 10.7 Design of Columns under an Eccentric Load 2
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.2 Stability of Structures 3 In the design of columns, cross-sectional area is selected such that - allowable stress is not exceeded - deformation falls within specifications After these design calculations, may discover that the column is unstable under loading and that it suddenly becomes sharply curved or buckles. all A P spec AE PL
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.2 Stability of Structures 4 Consider a model with two rods and torsional spring. After a small perturbation, Column is in stable equilibrium if If P > P cr then the column will continue to buckle. If P < P cr then the column will straighten up. moment ing destabiliz 2 L P sin 2 L P moment restoring 2 K 2 2 4 cr L P K K P P L
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.2 Stability of Structures 5 Assume that a load P is applied. After a perturbation, the system settles to a new equilibrium configuration at a finite deflection angle. Noting that sinθ < θ , the assumed configuration is only possible if P > P cr . 1.0152 1.0123 1.0097 1.0074 1.0054 1.0038 1.0024 1.0014 1.0006 1.0002 θ / sinθ 0.2955 0.2667 0.2377 0.2085 0.1790 0.1494 0.1197 0.0899 0.05996 0.029996 0 sin θ 0.3 0.27 0.24 0.21 0.18 0.15 0.12 0.09 0.06 0.03 0 θ sin 4 2 sin 2 cr P P K PL K L P 4 cr K P L 4 cr K P L
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.3 Euler’s Formula for Pin-Ended Beams 6 Consider an axially loaded beam. After a small perturbation, the system reaches an equilibrium configuration such that 0 2 2 2 2 y EI P dx y d y EI P EI M dx y d
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.3 Euler’s Formula for Pin-Ended Beams 7 Consider an axially loaded beam. After a small perturbation, the system reaches an equilibrium configuration such that Let Solution with be of the form y = A sin px + B cos px 0 2 2 2 2 y EI P dx y d y EI P EI M dx y d 2 2 2 0 d y p y dx 2 P p E I
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.3 Euler’s Formula for Pin-Ended Beams 8 now apply boundary conditions to determine A and B at x = 0, y = 0 at x= L, y = 0 0 = A sin (0) + B cos (0) 0 = A sin (pL) + B cos (pL) B = 0 A sin (pL) = 0 Either A=0 or sin(pL) = 0 pL = n π y = A sin px + B cos px 2 2 2 L I E n P 2 2 cr L I E P 2 2 P p E I P E I p n p L
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EGM 3520 Department of Mechanical and Aerospace Engineering 10.3 Euler’s Formula for Pin-Ended Beams 9 So, for the beam to have a shape of y = A sin ( px ) although we don’t have the value for A.
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  • Summer '08
  • DICKRELL
  • Trigraph, Buckling, Mechanical and Aerospace Engineering

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