1. 1 NASA Vehicle Assembly Building, 上一页下一页回目录 本章将介绍位移计算的功 - 能法，主要包括：功、应变能的基 本概念；功 - 能量方法计算位移（实功法）；桁架虚功法；梁和 框架虚功法；有限求和法；贝努利虚位移原理；麦克斯韦尔 - 贝 蒂位移互等原理。在前一章介绍梁和框架（包括刚架）结构位移 计算方法的基础上，进一步介绍更加通用的位移计算方法 —— 功 - 能法；将其拓展到虚功并提出虚功法 —— 虚功原理，进行位移 计算。以虚功原理为基础介绍位移互等定理，为后面章节的学习 提供基础。通过本章学习要求掌握挠度计算的功 - 能法，能够根 据结构的不同受力特征选用合适的方法进行挠度分析和计算；并 且掌握位移互等定理。 Chapter 10 Work-Energy Methods for Computing Deflections

2. 2 上一页下一页回目录 Chapter 10 Work — Energy Methods for Computing Deflections 10.1 Introduction 10.2 Work 10.3 Strain Energy 10.4 Deflections by the Work-Energy Method 10.5 Virtual Work: Trusses 10.6 Virtual Work: Beams and Frames 10.7 Finite Summation 10.8 Bernoulli’s Principle of Virtual Displacements 10.9 Maxwell-Betti Law of Reciprocal Deflections Summary

3. 3 When a structure is loaded, its stressed elements deform. As these deformations occur, the structure changes shape and points on the structure. In a well-designed structure, these displacements are small. 10.1 Introduction 上一页下一页回目录 As the truss shown in Figure 10.1, the applied load P produces axial forces F 1, F 2, and F 3 in the members. These forces cause the members to deform axially as shown by the dashed lines. As a result of these deformations, joint B of the truss displaces diagonally to B'.

4. 4 Work-energy methods provide the basis for several procedures used to calculate displacements. Work-energy lends itself to the computation of deflections because the unknown displacements can be incorporated directly into the expression for work-the product of a force and a displacement. 上一页下一页回目录 We will begin our study of work-energy by reviewing the work done by a force or moment moving through a small displacement. Then we will derive the equations for the energy stored in both an axially loaded bar and a beam. Finally, we will illustrate the work-energy method-also called the method of real work. 10.1 Introduction

5. 5 Since the method of real work has serious limitations i.e., deflections can be computed only at a point where a force acts and only a single concentrated load can be applied to the structure. The major emphasis in this chapter will be placed on the method of virtual work. It has many advantages. 上一页下一页回目录 Advantages applicable to many types of structural members virtual work permits the designer to include in deflection computations the influence of support settlements, temperature changes, creep, and fabrication errors. useful, versatile methods of computing deflections !!! The method does require that changes in geometry be small. 10.1 Introduction

6. 6 上一页下一页回目录 Work is defined as the product of a force times a displacement in the direction of the force. In deflection computations we will be concerned with the work done by both forces and moments. 10.2 Work If a force F and M remains constant,the work W may be expressed as follows:

7. 7 上一页下一页回目录 If a force varies in magnitude during a displacement and if the functional relationship between the force F and the collinear displacement d is known, the work can be evaluated by integration. 10.2 Work The displacement is divided into a series of small increments of length. The increment of work dW associated with each infinitesimal displacement equals F.

8. 8 上一页下一页回目录 The total work is then evaluated by summing all increments: 10.2 Work Similarly, for a variable moment that moves through a series of infinitesimal angular displacements, the total work is given as

9. 9 上一页下一页回目录 When a linear relationship exists between force and displacement and when the force increases from zero to its final value, expressions for work will always contain a one- half term, as shown by Equations 10.5 and 10.6 On the other hand, if the magnitude of a force or moment is constant during a displacement the work plots as a rectangular area and the one-half term is absent 10.2 Work

10. 10 10.3 Strain Energy 上一页下一页回目录 When a bar is loaded axially, it will deform and store strain energy U. For example, in the bar shown in Figure 10.4a, the externally applied load P induces an axial force F of equal magnitude (that is, F = P). 10.3.1 Truss Bars

11. 11 上一页下一页回目录 If the bar behaves elastically (Hooke ’ s law applies), the magnitude of the strain energy U stored in a bar by a force that increases linearly from zero to a final value F as the bar undergoes a change in length △ L equals. 10.3 Strain Energy 10.3.1 Truss Bars

12. 12 上一页下一页回目录 If the magnitude of the axial force remains constant as a bar undergoes a change in length △ L from some outside effect the strain energy stored in the member equals U = F ΔL Energy stored in a body, it also can be represented graphically like work by a force. If the variation of bar force is plotted against the change in bar length △ L, the area under the curve represents the strain energy U stored in the member. 10.3 Strain Energy 10.3.1 Truss Bars

13. 13 上一页下一页回目录 The figure is the graphic representation of Equation10.7 — the case in which a bar force increases linearly from zero to a final value F. The figure is the graphic representation of Equation 10.10 — the case in which the bar force remains constant as the bar changes length. 10.3 Strain Energy 10.3.1 Truss Bars

14. 14 上一页下一页回目录 10.3.2 beams 10.3 Strain Energy The increment of strain energy dU stored in a beam segment of infinitesimal length dx by a moment M that increases linearly from zero to a final value of M as the sides of the segment rotate through an angle equals.

15. 15 As we have shown previously, d θ may be expressed as To evaluate the total strain energy U stored in a beam of constant EI, the strain energy must be summed for all infinitesimal segments by integrating. 10.3.2 beams 10.3 Strain Energy

16. 16 If the moment M remains constant as a segment of beam undergoes a rotation d θ from another effect, the increment of strain energy stored in the element equals 10.3.2 beams 10.3 Strain Energy When dθ in Equation 10.14 is produced by a moment of magnitude M P, we can use Equation 9.13 to eliminate, dθand express dU as

17. 17 10.4 Deflections by the Work-Energy Method (Real Work) To establish an equation for computing the deflection of a point on a structure by the work-energy method, we can write according to the principle of conservation of energy that W = U where W is the work done by the external force applied to the structure and U is the strain energy stored in the stressed members of the structure. Equation 10.15 assumes that all work done by an external force is converted to strain energy.

18. 18 Because a single equation permits the solution of only one unknown variable, Equation 10.15 — the basis of the method of real work — can only be applied to structures that are loaded by a single force. Work-Energy Applied to a Truss To establish an equation that can be used to compute the deflection of a point on a truss due to a load P that increases linearly from zero to a final value P, we substitute Equations 10.5 and 10.9 into Equation 10.15 to give 10.4 Deflections by the Work-Energy Method (Real Work)

19. 19 As shown in Figure 10.5, joint B displaces both horizontally and vertically. Since the applied load of 30 kips is horizontal. Using the method of real work, determine the horizontal deflection δ x of joint B of the truss shown in Figure 10.5. For all bars, A=2.4 in 2 and E= 30,000 kips/in 2. The deflected shape is shown by the dashed lines. Example: 10.4 Deflections by the Work-Energy Method (Real Work)

20. 20 Since the applied force of P=30 kips acts in the direction of the required displacement, the method of real work is valid and Equation 10.16 applies. Values of bar force F are shown on the truss in Figure 10.5. 10.4 Deflections by the Work-Energy Method (Real Work) Solution :

21. 21 We are not able to compute the vertical component of the displacement of joint B by the method of real work because the applied force does not act in the vertical direction. The method of virtual work, which we discuss next, permits us to compute a single displacement component in any direction of any joint for any type of loading and thereby overcomes the major limitations of the method of real work. 10.4 Deflections by the Work-Energy Method (Real Work)

22. 22 Virtual work is a procedure for computing a single component of deflection at any point on a structure. To compute a component of deflection by the method of virtual work, the designer applies a force to the structure at the point and in the direction of the desired displacement. This force is often called a dummy load. The displacement it will undergo is produced by other effects. These other effects include the real loads, temperature change, support settlements, and so forth. 10.5 Virtual Work : Trusses 10.5.1 Virtual Work Method

23. 23 The dummy load and the reactions and internal forces it creates are termed a Q-system. Forces, work, displacements, energy associated with the Q system will be subscripted with a Q As to dummy load, typically we use a 1-kip or a 1-kN force to compute a linear displacement and a 1kipft or a 1 kN·m moment to determine a rotation or slope With the dummy load in place, the actual loads — called the P-system, are applied to the structure. Forces, deformations, work, and energy associated with the P-system will be subscripted with a P. 10.5 Virtual Work : Trusses 10.5.1 Virtual Work Method

24. 24 As the structure deforms under the actual loads, external virtual work W Q is done by the dummy load. In accordance with the principle of conservation of energy, an equivalent quantity of virtual strain energy U Q is stored in the structure; that is The virtual strain energy stored in the structure equals the product of the internal forces produced by the dummy load and the distortions of the elements of the structure produced by the real loads. 10.5 Virtual Work : Trusses 10.5.1 Virtual Work Method

25. 25 We will apply the method of virtual work to the one bar truss in Figure 10.6a to determine the horizontal displacement P of the roller at B. The bar, which carries axial load only, has a cross-sectional area A and modulus of elasticity E. As shows the bar force F P, the elongation of the bar L P, and the horizontal displacement δ P of joint B produced by the P system. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

26. 26 Since the bar is in tension, it elongates an amount △ L P, where Assuming that the horizontal load at joint B increases from zero to a final value P, we can use Equation 10.5 to express the real work W P done by force P as 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

27. 27 As a result of the real work done by P, strain energy U P of equal magnitude is stored in bar AB. Using Equation 10.7, we can express this strain energy as In accordance with the conservation of energy, W P equals U P, so the shaded areas W P and U P under the sloping lines in Figure 10.6b and c must be equal. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

28. 28 We next consider the work done on the strain energy stored in the bar by applying in sequence the dummy load Q followed by the real load P. We can express the real work W D done by the dummy load as: 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work Figure 10.6d shows the bar force F Q, the bar deformation △ L Q, and the horizontal displacement of joint B produced by the dummy load Q.

29. 29 The load-deflection curve associated with the dummy load is shown in Figure 10.6e. The triangular area under the sloping line represents the real work W D done by the dummy load Q. The corresponding strain energy U D stored in the bar as it elongates is equal to 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

30. 30 Figure 10.6f shows the strain energy stored in the structure due to the elongation of bar AB by the dummy load. In accordance with the principle of conservation of energy, W D must equal U D. Therefore, the crosshatched triangular areas in Figure 10.6e and f are equal. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

31. 31 Figure 10.6h shows the total work W t done by forces Q and P as point B displaces horizontally an amount δ t = δ Q + δ P. Figure 10.6i shows the total strain energy U t stored in the structure by the action of forces Q and P 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

32. 32 To clarify the physical significance of virtual work and virtual strain energy, we subdivide the areas in Figure 10.6h and i that represent the total work and total strain energy into the following three areas: 1. Triangular areas W D and U D (shown in vertical crosshatching). 2. Triangular areas W P and U P (shown in horizontal crosshatching). 3. Two rectangular areas labeled W Q and U Q. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

33. 33 Since W D =U D, W P =U P, and W t =U t by the principle of conservation of energy, it follows that the two rectangular areas W Q and U Q, which represent the external virtual work and the virtual strain energy, respectively, must be equal, and we can write As shown in Figure 10.6h, we can express W Q as 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

34. 34 where Q equals the magnitude of the dummy load and δ P the displacement or component of displacement in the direction of Q produced by the P-system. As indicated in Figure 10.6i, we can express U Q as where F Q is the bar force produced by the dummy load Q and ΔL P is the change in length of the bar produced by the P-system. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

35. 35 Substituting Equations 10.21a and 10.21b into Equation 10.17, we can write the virtual work equation for the one-bar truss as By adding summation signs to each side of Equation 10.22, we produce Equation 10.23, the general virtual work equation for the analysis of any type of truss. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

36. 36 The summation sign on the left side of Equation 10.23 indicates that in certain cases more than one external Q force contributes to the virtual work. The summation sign on the right side of Equation 10.23 is added because most trusses contain more than one bar. When the bar deformations are produced by load, we can use Equation 10.8 to express the bar deformations ΔL P in terms of the bar force F P and the properties of the members. For this case we can write Equation 10.23 as 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

37. 37 We will illustrate the use of Equation 10.24 by computing the deflection of joint B shown in example 10.2. First, we compute the component of displacement in the x direction, using a horizontal dummy load (shown in Fig. 10.7b). Then we compute the y component of displacement, using a vertical dummy load (shown in Fig. 10.7c). 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

38. 38 Example 10.2 Under the action of the 30- kip load, joint B of the truss in Figure 10.7a displaces to B (the deflected shape is shown by the dashed lines). Using virtual work, compute the components of displacement of joint B. For all bars, A=2 in 2 and E =30,000 kips/in 2. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

39. 39 To compute the horizontal displacement δ x of joint B, we apply a dummy load of 1 kip horizontally at B. Figure 10.7b shows the reactions and bar forces F Q produced by the dummy load. With the dummy load in place, we apply the real load of 30 kips to joint B (indicated by the dashed arrow). The 30-kip load produces bar forces F P, which deform the truss. For clarity we show the forces and deformations produced by the real load, P=30 kips, separately on the sketch in Figure 10.7a. Solution 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

40. 40 For clarity we show the forces and deformations produced by the real load, P=30 kips, separately on the sketch in Figure 10.7a. With the bar forces established, we use Equation 10.24 to compute δ x: 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work Solution

41. 41 To compute the vertical displacement δ y of joint B, we apply a dummy load of 1 kip vertically at joint B (see Fig. 10.7c) and then apply the real load. Since the value of F Q in bar AB is zero (see Fig. 10.7c), no energy is stored in that bar and we only have to evaluate the strain energy stored in bar BC. Using Equation 10.24, we compute 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work Solution

42. 42 The use of a 1-kip dummy load in Figure 10.7b and c was arbitrary, and the same results could have been achieved by applying a dummy force of any value. The direction of the dummy force may be selected arbitrarily, and the sign of the answer will automatically indicate the correct direction of the displacement. A positive sign signifies the displacement is in the direction of the dummy force; a negative sign indicates the displacement is opposite in sense to the direction of the dummy load. NOTE 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

43. 43 To evaluate the expression for virtual strain energy (F Q F P L)/(AE) on the right side of Equation 10.24 (particularly when a truss is composed of many bars), many engineers use a table to organize the computations (see Table 10.1 in Example 10.3). The total virtual strain energy stored in the truss equals the sum of the terms in column 6 divided by E. The value of the sum is written at the bottom of column 6. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work NOTE

44. 44 If E is a constant for all bars, it can be omitted from the summation and then introduced in the final step of the deflection computation If the value of either F Q or F P for any bar is zero, the strain energy in that bar is zero, and the bar can be omitted from the summation. 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work NOTE

45. 45 Compute the horizontal displacement δ x of joint B of the truss shown in Figure 10.8a. Given: E=30,000 kips/in 2, area of bars AD and BC=5 in 2 ; area of all other bars=4 in 2. Example 10.3 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

46. 46 Solution 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work

47. 47 Substituting ∑F Q F P L/AE =1025 into Equation 10.24 and multiplying the right side by 12 to convert feet to inches give 10.5 Virtual Work : Trusses 10.5.2 Analysis of Trusses by Virtual Work Solution

48. 48 As the temperature of a member varies, its length changes. An increase in temperature causes a member to expand; a decrease in temperature produces a contraction. In either case the change in length △ L temp can be expressed as where a = coefficient of thermal expansion, in/in per degree △ T = change in temperature L = length of bar 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

49. 49 To compute a component of joint deflection due to a change in temperature of a truss, first we apply a dummy load. Then we assume that the change in length of the bars produced by the temperature change occurs. external virtual work is done as the dummy load displaces. Internally, the change in length of the truss bars results in a change in strain energy U Q equal to the product of the bar forces F Q (produced by the dummy load) and the deformation △ L temp of the bars. 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

50. 50 A change in bar length L fabr due to a fabrication error is handled in exactly the same manner as a temperature change. If the bars of a truss change in length simultaneously due to load, temperature change, and a fabrication error, then △ L P in Equation 10.23 is equal to the sum of the various effects; that is The virtual work equation for computing a joint displacement can be established by substituting △ L temp for △ L P in Equation 10.23. 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

51. 51 When △ L P given by Equation 10.26 is substituted into Equation 10.23, the general form of the virtual work equation for trusses becomes Example 10.4 illustrates the computation of a component of truss displacement for both a temperature change and a fabrication error. 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

52. 52 Example 10.4 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error (1) bar BC fabricated 0.8 in too short and (2) bar AB fabricated 0.2 in too long. Given: a=6.5  10 6 in/in per. For the truss shown in Figure 10.9a, determine the horizontal displacement δx of joint B for a 60 increases in temperature and the following fabrication errors:

53. 53 Because the structure is determinate, no bar forces are created by either a temperature change or a fabrication error. If we imagine that the bars in their deformed state are connected to the pin supports at A and C (see Fig. 10.9c), bar AB will extend beyond point B a distance L AB to point c and the top of bar BC will be located a distance L BC below joint B at point a. Solution 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

54. 54 The deflected position of the truss is shown by the dashed lines. Since the initial displacement of each bar is directed tangent to the circle, we can assume for small displacements that the bars initially move in the direction of the tangent lines For example, as shown in Figure 10.9d in the region between points 1 and 2, the tangent line and the arc coincide closely. Solution 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

55. 55 Changes in length of bars due to temperature increase: To determine δx, we first apply a dummy load of 1kip at B (Fig. 10.9b) and then allow the specified bar deformations to take place. Using Equation 10.27, we compute Solution 10.5 Virtual Work : Trusses 10.5.3 Truss Deflections Produced by Temperature and Fabrication Error

56. 56 Structures founded on compressible soils often undergo significant settlements. These settlements can produce rotation of members and displacement of joints. If a structure is determinate, no internal stresses are created by a support movement. On the other hand, differential support settlements can induce large internal forces in indeterminate structures. The magnitude of these forces is a function of the member ’ s stiffness. 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

57. 57 Example 10.5 illustrates the use of virtual work to compute joint displacements and rotations produced by the settlements of the supports of a simple truss. The same procedure is applicable to determinate beams. Example 10.5 If support A of the truss in Figure 10.10a settles 0.6in and moves to the left 0.2in, determine (a) the horizontal displacement δx of joint B and (b) the rotation θ of bar BC. 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

58. 58 To compute δ x, apply a 1-kip dummy load horizontally at B (see Fig.10.10b) and compute all reactions. Assume that the support movements occur, evaluate the external virtual work, and equate to zero. (a) Solution 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

59. 59 Since no F P bar forces are produced by the support movement, F P = 0in Equation 10.24, yielding The minus sign indicates δ x is directed to the left.(b) To compute the rotation u of member BC, we apply a dummy load of 1kip. ft to bar BC anywhere between its ends and compute the support reactions (see Fig. 10.10c). 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements Solution

60. 60 The virtual work produced by a unit moment M Q used as a dummy load equals M Q θ. With this term added to W Q and with U Q =0, the expression for virtual work equals 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements Solution

61. 61 Compute the vertical displacement δy of joint C for the truss shown in Figure 10.11a. The truss bars are fabricated from an aluminum alloy whose stress-strain curve (see Fig. 10.11c) is valid for both uniaxial tension and compression. EXAMPLE 10.6 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

62. 62 The proportional limit, which occurs at a stress of 20 kips/in 2, divides elastic from inelastic behavior. Area of bar AC=1 in 2, and area of bar BC=0.5 in 2. In the elastic region E=10,000 kips/in 2. Solution The P-system with the F P forces noted on the bars is shown in Figure 10.11a. The Q-system with the F 0 forces is shown in Figure 10.11b 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

63. 63 To establish if bars behave elastically or are stressed into the inelastic region, we compute the axial stress and compare it to the proportional limit stress. For bar AC. Using Equation 10.8 gives For bar BC, Solution 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

64. 64 To compute ΔL P, we use Figure 10.11c to establish P. For σ=25 ksi, we read ε =0.008 in/in. Solution 10.5 Virtual Work : Trusses 10.5.4 Computation of Displacements Produced by Support Settlements

65. 65 For example, to compute the deflection at point C of the beam in Figure 10.14. Both shear and moment contribute to the deformations of beams we consider only deformations produced by moment 10.6 Virtual Work : Beams and Frames

66. 66 10.6 Virtual Work : Beams and Frames

67. 67 Using virtual work, compute (a) the deflection δ B and (b) the slope Ө B at the tip of the uniformly loaded cantilever beam in Figure 10.15a. EI is constant. Example 10.9 10.6 Virtual Work : Beams and Frames

68. 68 (a) compute the vertical deflection at B Solution 10.6 Virtual Work : Beams and Frames

69. 69 Solution 10.6 Virtual Work : Beams and Frames

70. 70 (b) To compute the slope at B, Solution 10.6 Virtual Work : Beams and Frames

71. 71 Considering the strain energy associated with both axial load and moment, compute the horizontal deflection of joint C of the frame in Figure 10.21a. Members are of constant cross section with I=600 in 4, A=13 in 2, and E=29,000 kips/in 2. Example 10.15 10.6 Virtual Work : Beams and Frames

72. 72 From A to B, x=0 to x=6 ft: From B to C, x=6 to x =15 ft: From D to C, x=0 to x=8 ft: Solution 10.6 Virtual Work : Beams and Frames

73. 73 Compute the horizontal displacement δ CH using virtual work. Consider both flexural and axial deformations in evaluating U Q. Only member AC carries axial load. Solution 10.6 Virtual Work : Beams and Frames

74. 74 If the depth or width of a member varies with distance along the member ’ s axis, the member is nonprismatic. If the functional relationship for the moment of inertia is complex, expressing it as a function of x may be difficult. In this situation, we can simplify the computation of the strain energy by replacing the integration (an infinitesimal summation) by a finite summation. 10.7 Finite Summation

75. 75 In a finite summation we divide a member into a series of segments, often of identical length. Properties are constant. We sum the contributions of all segments. We further simplify the summation by assuming that moments M Q and M P are constant over the length of the segment and equal to the values at the center of the segment. 10.7 Finite Summation

76. 76 We can represent the virtual strain energy in a finite summation by the following equation: Although a finite summation produces an approximate value of strain energy, the accuracy of the result is usually good even when a small number of segments (say, five or six) are used. 10.7 Finite Summation

77. 77 Using a finite summation, compute the deflection δ B of the tip of the cantilever beam in Figure 10.22a. The 12-in-wide beam has a uniform taper, and E=3000 kips/in 2. 10.7 Finite Summation EXAMPLE 10.16

78. 78 Values of M Q and M P are tabulated in columns 4 and 5 of Table 10.3. Using Equation 10.34 to evaluate the right side of Equation 10.31, solve for δ B. 10.7 Finite Summation Solution Divide the beam length into four segments

79. 79 10.7 Finite Summation Solution

80. 80 Bernoulli ’ s principle of virtual displacements, is a variation of the principle of virtual work. It can be used to compute the deflection of points on a determinate structure that undergoes rigid body movement. In Bernoulli ’ s principle, virtual work equals the product of each force or moment and the component of the virtual displacement through which is moved. Thus it can be expressed by the equation. 10.8 Bernoulli’s Principle of Virtual Displacements

81. 81 The external effect of a system of forces acting on a body can always be replaced by a resultant force R through any point and a moment M. 10.8 Bernoulli’s Principle of Virtual Displacements Since the body is rigid, U Q = 0.

82. 82 Linear displacement △ L has components △ x in the x direction and △ y in the y direction, the virtual work W Q produced by these displacements equals Since Equation 10.36 establishes 10.8 Bernoulli’s Principle of Virtual Displacements Or by expressing R in terms of its rectangular components,

83. 83 If support B of the L-shaped beam in Figure 10.23a settles 1.2in, determine (a) the vertical displacements δ C of point C, (b) the horizontal displacement δ D of point D, and (c) the slope θ A at point A. 10.8 Bernoulli’s Principle of Virtual Displacements EXAMPLE 10.17

84. 84 We next compute the reactions at the supports using the equations of statics. In accordance with Bernoulli ’ s principle, we equate to zero the sum of the virtual work done by the Q system forces. To compute the vertical displacement at C, we apply a 1-kip dummy load in the vertical direction at C (see Fig. 10.23b). (a) 10.8 Bernoulli’s Principle of Virtual Displacements Solution

85. 85 (b) To compute the horizontal displacement of joint D See Fig. 10.23c then compute the virtual work and set it equal to zero. 10.8 Bernoulli’s Principle of Virtual Displacements

86. 86 (c) To compute θ A 10.8 Bernoulli’s Principle of Virtual Displacements

87. 87 10.9 Maxwell-Betti Law of Reciprocal Deflections The Maxwell-Betti law, states: A linear deflection component at a point A in direction 1 produced by the application of a unit load at a second point B in direction 2 is equal in magnitude to the linear deflection component at point B in direction 2 produced by a unit load applied at A in direction 1. solve indeterminate structures of two or more degrees of indeterminacy the construction of indeterminate influence lines. any stable elastic structure applications

88. 88 We can establish Maxwell ’ s law by considering the deflections at points A and B of the beam in Figure 10.25a and b. Displacements are labeled with two subscripts. The first subscript indicates the location of the displacement. The second subscript indicates the point at which the load producing the displacement acts. 10.9 Maxwell-Betti Law of Reciprocal Deflections

89. 89 We evaluate the total work done by the two forces F A and F B when they are applied in different order to the simply supported beam. The forces are assumed to increase linearly from zero to their final value. Since the final deflected position of the beam produced by the two loads is the same, the total work done by the forces is also the same. 10.9 Maxwell-Betti Law of Reciprocal Deflections

90. 90 Case 1. F B Applied Followed by F A (a) Work done when F B is applied: (b) Work done when F A is applied with F B in place: Since the magnitude of F B does not change as the beam deflects under the action of F A, the additional work done by F B (the second term in the equation above) equals the full value of F B times the deflection △ BA produced by F A. 10.9 Maxwell-Betti Law of Reciprocal Deflections

91. 91 Case 2. F A Applied Followed by F B (c) Work done when F A is applied: (d) Work done when F B is applied with F A in place: 10.9 Maxwell-Betti Law of Reciprocal Deflections

92. 92 10.9 Maxwell-Betti Law of Reciprocal Deflections Equating the total work of cases 1 and 2 When F A and F B 1 kip,

93. 93 As for rotations,we state the Maxwell-Betti: The rotation at point A in direction 1 due to a unit couple at B in direction 2 is equal to the rotation at B in direction 2 due to a unit couple at A in direction 1. As a third variation of the Maxwell-Betti law, we can also state: Any linear component of deflection at a point A in direction 1 produced by a unit moment at B in direction 2 is equal in magnitude to the rotation at B (in radians) in direction 2 due to a unit load at A in direction 1. 10.9 Maxwell-Betti Law of Reciprocal Deflections

94. 94 10.9 Maxwell-Betti Law of Reciprocal Deflections The figure also shows that AB is the same direction as the load at A, and the rotation and the moment at B are in the same counterclockwise direction. The rotation at point B produced by the unit load at A is equal in magnitude to the vertical deflection △ AB at A produced by the unit moment at point B.

95. 95 In its most general form, the Maxwell-Betti law can also be applied to a structure that is supported in two different ways. It may be represented by the following equation: where F 1 represents a force or moment in system 1 and δ 2 is the displacement in system 2 that corresponds to F 1. Similarly, F 2 represents a force or moment in system 2, and δ 1 is the displacement in system 1 that corresponds to F 2. 10.9 Maxwell-Betti Law of Reciprocal Deflections

96. 96 Figure 10.28 shows the same beam supported and loaded in two different ways. Demonstrate the validity of Equation 10.41. Required displacements are noted on the figure. 10.9 Maxwell-Betti Law of Reciprocal Deflections EXAMPLE 10.18

97. 97 Summary Virtual work permits the engineer to compute a single component of deflection with each application of the method. Virtual work assumes loads are applied slowly so that neither kinetic nor heat energy is produced. 10.9 Maxwell-Betti Law of Reciprocal Deflections Solution

98. 98 External virtual work W Q is done by the dummy loads. Simultaneously an equivalent quantity of virtual strain energy U Q is stored in the structure. That is, We limit the application of the method to three of the most common types of planar structures: trusses, beams, and frames, neglect the effects of shear. If a deflection has both vertical and horizontal components, The actual deflection is the vector sum of the two orthogonal components Summary

99. 99 Summary The use of a unit load to establish a Q system is arbitrary. To determine the virtual strain energy when the depth of a beam varies along its length, changes in cross-sectional properties can be taken into account by dividing the beam into segments and carrying out a finite summation (see Sec. 10.7). In Section 10.9, we introduce the Maxwell-Betti law of reciprocal deflections. This law will be useful when we set up the terms of the symmetric matrices required to solve indeterminant structures.

100. 100 Summary 总结 虚功法是第 10 章的主题。基于能量守恒原理，虚功法假设荷载是缓 慢施加的，因此既不产生动能也不产生热能 。 ，同时结构中产生大小相等的应变能 ，即 用虚功法计算某点位移时，我们假设一个力（称为虚力）作用在该 点所求位移的方向上，这个力和它的反力称为 Q 力系。如果要求的 是转角，则施加虚力矩。在虚力作用的同时，实际荷载 —— 称为 P 力系 —— 也作用在结构上。当结构在实际荷载作用下产生变形时， 虚力在 P 力系引起的实际位移上做虚功 ，同时结构中产生大小 相等的应变能 ，即， = 这里我们将虚功法的应用限定在三种最普通的平面结构，即桁架、 梁和框架。我们还忽略剪力的影响，只有对较短的承载很大的深梁 或者刚性模量较小的梁，剪力的影响较大。该方法还可以用来求解 由温度变化、支座沉降以及制造误差引起的位移。

101. 101 Summary 如果位移既有水平分量又有竖直分量，就要分两次应用虚功法求解。 不一定要使用单位荷载来建立 Q 力系。但是，由于单位荷载引起的 位移（称为柔度系数）在超静定结构分析中需要用到（见第 11 章），因此实际应用中通常都采用单位荷载。 要计算变截面梁的虚应变能，可以采用有限求和法，将梁划分为若 干小段，以考虑横截面特性的变化 。 第 10.9 节介绍了麦克斯韦尔 —— 贝蒂位移互等定理，这个定理在第 11 章中采用柔度法建立求解超静定结构时所需的对称矩阵时非常 有用。