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Chapter 11

Chapter 11. Equilibrium and Elasticity. Equilibrium. Two Conditions for Equilibrium. To motivate these, recall:. Defining Equilibrium. Equilibrium = no net external force or torque = no change in translation or rotation) your text says L =0; others allow nonzero L :.

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Chapter 11

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  1. Chapter 11 Equilibrium and Elasticity

  2. Equilibrium

  3. Two Conditions for Equilibrium • To motivate these, recall:

  4. Defining Equilibrium • Equilibrium= no net external force or torque = no change in translation or rotation) • your text says L=0; others allow nonzero L:

  5. Defining Static Equilibrium • ‘Static’ Equilibrium= the special case of no translation or rotation at all

  6. Two Conditions for Equilibrium • When applying these, we must consider all external forces • But the gravitational force is rather subtle

  7. Center of Gravity (cg) • Gravity acts at every point of a body • Let t = the torque on a body due to gravity • Can find t by treating the body as a single particle (the ‘cg’)

  8. Center of Mass (cm) • it can be shown: if g = constant everywhere, then: • center of gravity =center of mass

  9. Using the Center of Gravity Pressent some more explanatory notes

  10. SolvingEquilibrium Problems

  11. Two Conditions for Equilibrium • From now on, in this chapter/lecture: • center of mass = center of gravity • ‘equilibrium’ means ‘static equilibrium’ • write: SF and St for SFext and Stext

  12. First Conditionfor Equilibrium

  13. Second Condition for Equilibrium

  14. Exercise 11-11 Work through Exercise 11-11

  15. Exercise 11-14 Work through Exercise 11-14

  16. A different version of Example 11-3 The ‘Leaning Ladder’ Problem Work through the variation the the text’s leaning ladder problem

  17. Problem 11-62 ‘Wheel on the Curb’ Problem Work through Problem 11-62

  18. Elasticity

  19. Elasticity • Real bodies are not perfectly rigid • They deform when forces are applied • Elastic deformation: body returns to its original shape after the applied forces are removed

  20. Stress and Strain • stress: describes the applied forces • strain: describes the resulting deformation • Hooke’s Law: stress = modulus × strain • modulus: property of material under stress • (large modulus means small deformation)

  21. Hooke’s Law and Beyond • O to a : • small stress, strain • Hooke’s Law:stress=modulus×strain • a < b : • stress and strain areno longer proportional

  22. Units • stress = modulus × strain • stress (‘applied force’): pascal= Pa=N/m2 • strain (‘deformation’): dimensionless • modulus: same unit as stress

  23. Types of Stress and Strain • Applied forces are perpendicular to surface: • tensile stress • bulk (volume) stress • Applied forces are parallel to surface: • shear stress

  24. Tensile Stress and Strain • tensile stress = F/A • tensile strain = Dl/l0 • Young’s modulus = Y

  25. Tensile Stress and Strain Work through Exercise 11-22

  26. Compression vs. Tension • tension (shown): pull on object • compression: push on object(reverse directionof F shown at left) • Ycompressive = Ytensile Work through Exercise 11-26

  27. Tension and Compression at once

  28. Bulk Stress and Strain • pressure: p=F/A • bulk stress = Dp • bulk strain = DV/V0 • bulk modulus = B

  29. Bulk Stress and Strain • B > 0 • negative sign above:Dp and DV have opposite signs Work through Exercise 11-30

  30. Shear Stress and Strain

  31. Shear Stress and Strain • shear stress = F7/A • shear strain = x/h = tanf • shear modulus = S

  32. Shear Stress and Strain Do Exercise 11-32

  33. Regimes of Deformation • O to a : • (small stress, strain) • stress=modulus×strain • elastic, reversible • a < b : • elastic, reversible • but stress and strainnot proportional

  34. Regimes of Deformation • From point O to b : • elastic, reversible • from point b to d: • plastic, irreversible • ductile materials havelong c–d curves • brittle materials haveshort c–d curves

  35. Demonstation Tensile Strength and Fracture

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