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The following will be discussed in this chapter: Strain energy, strain energy density, elastic strain energy for normal stresses, strain energy for shearing stresses, strain energy for a general state of stress, design for impact loads, work and energy under a single load,...
Third Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P Beer E Russell Johnston, Jr John T DeWolf Energy Methods Lecture Notes: J Walt Oler Texas Tech University © 2002 The McGraw-Hill Companies, Inc All rights reserved Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Energy Methods Strain Energy Strain Energy Density Elastic Strain Energy for Normal Stresses Strain Energy For Shearing Stresses Sample Problem 11.2 Strain Energy for a General State of Stress Impact Loading Example 11.06 Example 11.07 Design for Impact Loads Work and Energy Under a Single Load Deflection Under a Single Load © 2002 The McGraw-Hill Companies, Inc All rights reserved Sample Problem 11.4 Work and Energy Under Several Loads Castigliano’s Theorem Deflections by Castigliano’s Theorem Sample Problem 11.5 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain Energy • A uniform rod is subjected to a slowly increasing load • The elementary work done by the load P as the rod elongates by a small dx is dU = P dx = elementary work which is equal to the area of width dx under the loaddeformation diagram • The total work done by the load for a deformation x1, x1 U = ∫ P dx = total work = strain energy which results in an increase of strain energy in the rod • In the case of a linear elastic deformation, x1 U = ∫ kx dx = 12 kx12 = 12 P1x1 © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain Energy Density • To eliminate the effects of size, evaluate the strainenergy per unit volume, U = V x1 P dx ∫A L ε1 u = ∫ σ x dε = strain energy density • The total strain energy density resulting from the deformation is equal to the area under the curve to ε1 • As the material is unloaded, the stress returns to zero but there is a permanent deformation Only the strain energy represented by the triangular area is recovered • Remainder of the energy spent in deforming the material is dissipated as heat © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain-Energy Density • The strain energy density resulting from setting ε1 = εR is the modulus of toughness • The energy per unit volume required to cause the material to rupture is related to its ductility as well as its ultimate strength • If the stress remains within the proportional limit, ε1 Eε12 σ 12 = u = ∫ Eε1 dε x = 2E • The strain energy density resulting from setting σ1 = σY is the modulus of resilience uY = © 2002 The McGraw-Hill Companies, Inc All rights reserved σ Y2 2E = modulus of resilience 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Elastic Strain Energy for Normal Stresses • In an element with a nonuniform stress distribution, ∆U dU = V ∆ dV ∆V →0 u = lim U = ∫ u dV = total strain energy • For values of u < uY , i.e., below the proportional limit, σ x2 U =∫ 2E dV = elastic strain energy • Under axial loading, σ x = P A dV = A dx L P2 U =∫ dx AE • For a rod of uniform cross-section, P2L U= AE © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Elastic Strain Energy for Normal Stresses • For a beam subjected to a bending load, σ x2 M y2 U =∫ 2E dV = ∫ EI dV • Setting dV = dA dx, M ⎛⎜ ⎞⎟ U=∫ ∫ dA dx = ∫ y dA dx 2⎜∫ ⎟ EI EI ⎝ A ⎠ 0 A L σx = My I L =∫ M y2 L M2 dx EI • For an end-loaded cantilever beam, M = − Px L P2 x2 P L3 U=∫ dx = EI EI © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain Energy For Shearing Stresses • For a material subjected to plane shearing stresses, γ xy u= ∫τ xy dγ xy • For values of τxy within the proportional limit, u= Gγ xy = 12 τ xy γ xy = τ xy 2G • The total strain energy is found from U = ∫ u dV =∫ © 2002 The McGraw-Hill Companies, Inc All rights reserved τ xy 2G dV 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain Energy For Shearing Stresses • For a shaft subjected to a torsional load, τ xy T 2ρ U =∫ 2G dV = ∫ 2GJ dV • Setting dV = dA dx, T ⎛⎜ ⎞⎟ U =∫∫ dA dx = ∫ ρ dA dx 2⎜∫ ⎟ 2GJ 2GJ ⎝ A ⎠ 0A L τ xy = Tρ J T 2ρ L L T2 =∫ dx 2GJ • In the case of a uniform shaft, T 2L U= 2GJ © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.2 SOLUTION: • Determine the reactions at A and B from a free-body diagram of the complete beam • Develop a diagram of the bending moment distribution a) Taking into account only the normal stresses due to bending, determine the strain energy of the beam for the loading shown b) Evaluate the strain energy knowing that the beam is a W10x45, P = 40 kips, L = 12 ft, a = ft, b = ft, and E = 29x106 psi © 2002 The McGraw-Hill Companies, Inc All rights reserved • Integrate over the volume of the beam to find the strain energy • Apply the particular given conditions to evaluate the strain energy 11 - 10 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 11.07 SOLUTION: • The normal stress varies linearly along the length of the beam as across a transverse section • Find the static load Pm which produces the same strain energy as the impact • Evaluate the maximum stress A block of weight W is dropped from a resulting from the static load Pm height h onto the free end of the cantilever beam Determine the maximum value of the stresses in the beam © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - 17 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Example 11.07 • Find the static load Pm which produces the same strain energy as the impact For an end-loaded cantilever beam, Pm2 L3 Um = EI Pm = SOLUTION: • The normal stress varies linearly along the length of the beam as across a transverse section =∫ 2E dV ≠ L3 • Evaluate the maximum stress resulting from the static load Pm U m = Wh σ m2 6U m EI σ m2 V 2E © 2002 The McGraw-Hill Companies, Inc All rights reserved M m c Pm Lc = σm = I I = 6U m E ( ) = LI c 6WhE ( ) L I c2 11 - 18 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Design for Impact Loads • For the case of a uniform rod, σm = 2U m E V • For the case of the nonuniform rod, σm = 16 U m E AL V = A( L / ) + A( L / ) = AL / σm = 8U m E V • For the case of the cantilever beam Maximum stress reduced by: • uniformity of stress • low modulus of elasticity with high yield strength high volume â 2002 The McGraw-Hill Companies, Inc All rights reserved σm = 6U m E ( ) L (I / c ) = L (14 πc / c ) = 14 (πc L ) = 14 V σm = L I c2 24U m E V 11 - 19 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Work and Energy Under a Single Load • Strain energy may also be found from the work of the single load P1, x1 U = ∫ P dx • For an elastic deformation, • Previously, we found the strain energy by integrating the energy density over the volume For a uniform rod, U = ∫ u dV = ∫ L =∫ σ 2E dV (P1 A)2 Adx = 2E x1 x1 0 U = ∫ P dx = ∫ kx dx = 12 k x12 = 12 P1x1 • Knowing the relationship between force and displacement, P12 L AE © 2002 The McGraw-Hill Companies, Inc All rights reserved PL x1 = AE P L P L ⎛ ⎞ 1 U = 12 P1⎜ = ⎟ ⎝ AE ⎠ AE 11 - 20 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Work and Energy Under a Single Load • Strain energy may be found from the work of other types of single concentrated loads • Transverse load U= y1 1Py P dy = ∫ 1 ⎛ P1L3 ⎞ ⎟= = P1⎜ ⎜ 3EI ⎟ ⎝ ⎠ • Bending couple θ1 U = ∫ M dθ = 12 M1θ1 P12 L3 EI M L M L ⎛ ⎞ = 12 M1⎜ ⎟ = ⎝ EI ⎠ EI © 2002 The McGraw-Hill Companies, Inc All rights reserved • Torsional couple φ1 U = ∫ T dφ = 12 T1φ1 T L T L ⎛ ⎞ = 12 T1⎜ ⎟ = ⎝ JG ⎠ JG 11 - 21 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Deflection Under a Single Load • If the strain energy of a structure due to a single concentrated load is known, then the equality between the work of the load and energy may be used to find the deflection • Strain energy of the structure, 2 FBC LBC FBD LBD + U= AE AE [ From the given geometry, LBC = 0.6 l LBD = 0.8 l From statics, FBC = +0.6 P FBD = −0.8 P ] P 2l (0.6 )3 + (0.8)3 P 2l = = 0.364 AE AE • Equating work and strain energy, P2L U = 0.364 = P yB AE yB = 0.728 © 2002 The McGraw-Hill Companies, Inc All rights reserved Pl AE 11 - 22 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.4 SOLUTION: • Find the reactions at A and B from a free-body diagram of the entire truss • Apply the method of joints to determine the axial force in each member Members of the truss shown consist of sections of aluminum pipe with the cross-sectional areas indicated Using E = 73 GPa, determine the vertical deflection of the point E caused by the load P © 2002 The McGraw-Hill Companies, Inc All rights reserved • Evaluate the strain energy of the truss due to the load P • Equate the strain energy to the work of P and solve for the displacement 11 - 23 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.4 SOLUTION: • Find the reactions at A and B from a freebody diagram of the entire truss Ax = −21 P Ay = P B = 21 P • Apply the method of joints to determine the axial force in each member FDE = − 17 P FAC = + 15 P FDE = 54 P FCE = + 15 P FCD = FCE = − 21 P © 2002 The McGraw-Hill Companies, Inc All rights reserved FAB = 11 - 24 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.4 • Evaluate the strain energy of the truss due to the load P Fi2 Li Fi2 Li U =∑ = ∑ Ai Ai E E = ( 29700 P 2E ) • Equate the strain energy to the work by P and solve for the displacement Py = U E 2U ⎛⎜ 29700 P ⎞⎟ yE = = P P ⎜⎝ E ⎟⎠ ( 29.7 × 103 )(40 × 103 ) yE = 73 ì 10 â 2002 The McGraw-Hill Companies, Inc All rights reserved y E = 16.27 mm ↓ 11 - 25 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Work and Energy Under Several Loads • Deflections of an elastic beam subjected to two concentrated loads, x1 = x11 + x12 = α11P1 + α12 P2 x2 = x21 + x22 = α 21P1 + α 22 P2 • Compute the strain energy in the beam by evaluating the work done by slowly applying P1 followed by P2, ( U = 12 α11P12 + 2α12 P1P2 + α 22 P22 ) • Reversing the application sequence yields ( U = 12 α 22 P22 + 2α 21P2 P1 + α11P12 ) • Strain energy expressions must be equivalent It follows that α12=α21 (Maxwell’s reciprocal theorem) © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - 26 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Castigliano’s Theorem • Strain energy for any elastic structure subjected to two concentrated loads, ( U = 12 α11P12 + 2α12 P1P2 + α 22 P22 ) • Differentiating with respect to the loads, ∂U = α11P1 + α12 P2 = x1 ∂P1 ∂U = α12 P1 + α 22 P2 = x2 ∂P2 • Castigliano’s theorem: For an elastic structure subjected to n loads, the deflection xj of the point of application of Pj can be expressed as xj = © 2002 The McGraw-Hill Companies, Inc All rights reserved ∂U ∂Pj and θ j = ∂U ∂M j φj = ∂U ∂T j 11 - 27 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Deflections by Castigliano’s Theorem • Application of Castigliano’s theorem is simplified if the differentiation with respect to the load Pj is performed before the integration or summation to obtain the strain energy U • In the case of a beam, L M2 U=∫ dx EI L M ∂M ∂U xj = dx =∫ EI ∂Pj ∂Pj • For a truss, n Fi2 Li U =∑ 2A E i =1 i © 2002 The McGraw-Hill Companies, Inc All rights reserved n ∂U F L ∂F =∑ i i i xj = ∂Pj i =1 Ai E ∂Pj 11 - 28 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.5 SOLUTION: • For application of Castigliano’s theorem, introduce a dummy vertical load Q at C Find the reactions at A and B due to the dummy load from a free-body diagram of the entire truss Members of the truss shown consist of sections of aluminum pipe with the cross-sectional areas indicated Using E = 73 GPa, determine the vertical deflection of the joint C caused by the load P • Apply the method of joints to determine the axial force in each member due to Q • Combine with the results of Sample Problem 11.4 to evaluate the derivative with respect to Q of the strain energy of the truss due to the loads P and Q • Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - 29 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.5 SOLUTION: • Find the reactions at A and B due to a dummy load Q at C from a free-body diagram of the entire truss Ax = − 34 Q Ay = Q B = 34 Q • Apply the method of joints to determine the axial force in each member due to Q FCE = FDE = FAC = 0; FCD = −Q FAB = 0; FBD = − 34 Q © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - 30 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Sample Problem 11.5 • Combine with the results of Sample Problem 11.4 to evaluate the derivative with respect to Q of the strain energy of the truss due to the loads P and Q ⎛ F L ⎞ ∂F yC = ∑ ⎜⎜ i i ⎟⎟ i = (4306 P + 4263Q ) ⎝ Ai E ⎠ ∂Q E • Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C yC = ( 4306 40 × 103 N 73 × 10 Pa ) yC = 2.36 mm ↓ © 2002 The McGraw-Hill Companies, Inc All rights reserved 11 - 31 ...Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Energy Methods Strain Energy Strain Energy Density Elastic Strain Energy for Normal Stresses Strain Energy For Shearing Stresses... reserved 11 - Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain -Energy Density • The strain energy density resulting from setting ε1 = εR is the modulus of toughness • The energy. .. McGraw-Hill Companies, Inc All rights reserved 11 - 12 Third Edition MECHANICS OF MATERIALS Beer • Johnston • DeWolf Strain Energy for a General State of Stress • Previously found strain energy