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Chapter 11 Energy Method 

Chapter 11 Energy Method . -- Utilize the Energy Method to solve engineering mechanics problems. -- Set aside the Equations of quilibrium. 1. Introduction. The relations between forces and deformation :. Stress -- Ch 1. Fundamental concept of. Strain – Ch 2.

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Chapter 11 Energy Method 

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  1. Chapter 11 Energy Method  -- Utilize the Energy Method to solve engineering mechanics problems. -- Set aside the Equations of quilibrium

  2. 1. Introduction The relations between forces and deformation : Stress -- Ch 1 Fundamental concept of Strain – Ch 2 Strain Energy – Ch 11 We will learn: • Modulus of Toughness • Modulus of resilience • 3. Castigliano Theorem

  3. 11.2 Strain Energy (11.1) (11.2)

  4. If the material response is elastic: (11.3)

  5. 11.3 Strain-Energy Density (11.4) Modulus of Toughness = Toughness = area under the - curve.

  6. (11.5) (11.6) (11.7) (11.8) Modulus of Resilience:

  7. 11.4 Elastic strain Energy for Normal Stresses (11.9)

  8. (11.9) (11.11) (11.12)

  9. (11.13)

  10. 11.5 Elastic Strain Energy for Shearing Stresses (11.18) (11.19) (11.20) (11.21)

  11. Strain Energy in Torsion (11.19) (11.21) (11.22)

  12. 11.6 Strain Energy for a General State of Stress (11.25) From Eq. (2.38) (2.38) (11.26)

  13. (11.26) If the principal stresses are used: (11.27) Where a, b, c = the principal stresses

  14. (11.28) Where uv = the part of energy leading to volume change = hydrostatic stress ud = deviatoric energy = the part of energy leading to shape change. Defining Mean Stress or Average Stress (11.29)

  15. And set: (11.30) where = mean stress = deviatoric stress Combining Eqs. (11-29) and (11.30) (11.31)

  16. (11.31) -- They only change the shape, but do not lead to the change of the volume of the material. The dilatation, V/V, caused by the state of stress can be obtained, via Eq. (2.31) as or e = 0

  17. The uv can be obtained by substituting into Eq. (11.27) to obtain By means of Eq. (11.29) we have (11.32)

  18. The distortion or deviatoric energy can be obtained as After simplification: Recalling Eq. (2.43) or (2.43)

  19. Hence, the previous equation takes a new form (11.33) For 2-D cases, Eq. (11.33) reduces to (11.34) For uniaxial tension, i.e. 1-D cases,b = 0 and a = y

  20. Substituting this equation to the previous equation, it leads to (7.26) Expanding the same operation to a 3-D case, one can have (11.35) Replacing “<“ by “=“, it follows (11.36) This is a circular cylinder of radius

  21. The 2- D Yield Locus: (7.26)

  22. The 3- D Yield Locus: (11.35)

  23. 11.7 Impact Loading K. E. of the ball = K. E. = The strain energy in the bar: (11.37) Assuming: 1. No heat dissipation 2. The ball sticks with the rod after impact. (11.38)

  24. If the stress is uniform within the rod: Therefore, m can be determined as: (11.39)

  25. 11.8 Design for Impact Loads Case A: For a Uniform-Diameter Rod: (11.45a) Case B: For a Multiple-Section Rod: (11.45b)

  26. Case C: For a Circular Cantilever Rod: (11.44) However, Hence, Eq. (11.44) can be reduced to (11.45c)

  27. Since We conclude that: --- in order to develop lower m in the rod, the rod should have 1. Lower E 2. Larger V 3. Uniform m

  28. 11.9 Work and Energy under a Single Load Case A: For an Uniaxial Load: (11.2) (11.3) Case B: For a Cantilever Beam: (11.46)

  29. Case C: For a Beam in Bending: (11.47) (11.48) Case D: For a Beam in Torsion: (11.49)

  30. 11.10 Deflection under a Single Load by the Work-Energy Method Case A: For an Uniaxial Load: (11.3) where x1 = deflection due to P1 Case B: For a Beam in Bending: (11.47) where 1 = deflection due to single moment M1

  31. 11.11 Work and Energy under Several Loads Deflection due to P1: (11.54) Deflection due to P2: (11.54) ij = influence coefficients The combining effect of P1 and P2 (11.54) (11.54)

  32. Calculating the Work Done by P1 and P2: Case I: Assuming P1 is applied first --- At Point 1, the work done by P1 is (11.58) At Point 2, the work done by P1 is zero. P2 is applied next --- At Point 2, the work done by P2 is (11.59)

  33. At Point 1, the work done by P1 due to additional defection [caused by P2] is (11.60) (11.58) + (11.59) + (11.60), the total strain energy is (11.61)

  34. Case II: Assuming P2 is applied first --- (11.62) Equating Eqs. (11.61) and (11.62) leads to -- Maxwell Reciprocal Theorem

  35. 11.12 Castigliano’s Theorem (11.61) (11.63) (11.64) Or, in general [ Castigliano’s Theorem ] (11.65) (11.66)

  36. General Formulation for Castigliano’s Theorem For multiple loading, P1, P2, …., Pn the deflection of the point of application of Pi can be expressed as (11.66) The total strain energy of the structure is (11.67) Differentiating U w.r.to Pj

  37. Since ij = ji, the above equation becomes (11.65) -- for concentrated loads -- for moment loads -- for torsion

  38. 11.13 Deflections by Castigliano’s Theorem Total strain energy of a beam subjected to bending But, the differentiation can be applied prior to integration. (11.17) Deflection at point Pj (11.70) Total strain energy of a truss member (11.71) Deflection at point Pj (11.72)

  39. If no load is applied to a point, where we desire to obtain a deflection: -- Apply a dummy (fictitious) load Q at that point, determine [ Castigliano’s Theoem ] (11.76) Then, set Q = 0.

  40. 11.14 Statically Indeterminate Structures Structure indeterminate to the 1st degree: Procedures: 1. Assume one support as redundant 2. Replace it with an unknown force 3. y = U/RA = 0  solving for RA

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