1 Outerbridge Crossing 本章学习桁架的相关内容，主要包括桁架的构成特点、桁 架的种类、桁架分析的基本思路、节点法、零杆、截面法、静 定性和稳定性以及桁架的计算机分析。典型桁架结构由仅承受 轴力的杆件组成，分析比较简单，关键在于能够熟练灵活地应 用隔离体图和静力平衡条件。当桁架构成复杂结构时，注意发 现其构成特点和规律对简化分析、保证分析的正确性十分有益。 桁架分析中，尽可能找出零杆常常可以简化分析；结点法是基 本的分析方法，可用于任意情况下桁架的分析；但对复杂桁架， 采用巧妙的截面将某一部分截开形成隔离体并应用平衡条件， 常可大大简化分析，对只需求解个别杆件内力的情形更是如此。 本章是结构力学的基本部分，必须熟练掌握。 CHAPTER 4 Trusses

2 chapter 4 Trusses 4.1 Introduction 4.2 Types of Trusses 4.3 Analysis of Trusses 4.4 Method of Joints 4.5 Zero Bars 4.6 Method of Sections 4.7 Determinacy and Stability 4.8 Computer Analysis of Trusses Summary 上一页下一页回目录

3 上一页下一页回目录 4.1 Introduction A truss is a structural element composed of a stable arrangement of slender interconnected bars The pattern of bars is selected to produce an efficient, lightweight, load-bearing member.

4 上一页下一页回目录 Although joints are rigid, the designer normally assumes that members are connected at joints by frictionless pins and truss members are assumed to carry only axial force As a matter of fact, a truss can often be viewed as a beam in which excess material has been removed to reduce weight. The primary function of the vertical and diagonal members is to transfer vertical force (shear) to the supports at the ends of the truss. 4.1 Introduction the truss will require less material because the material is used more efficiently

5 上一页下一页回目录 4.1 Introduction the designer can vary the truss depth to reduce its weight

6 上一页下一页回目录 4.1 Introduction The diagonals of a truss typically slope upward at an angle that ranges from 45°to 60°. In a long-span truss the distance between panel points should not exceed 15 to 20 ft (5 to 7 m) to limit the unsupported length of the compression chords, which must be designed as columns. As the slenderness of a compression chord increases, it becomes more susceptible to buckling. The slenderness of tension members must be limited also to reduce vibrations produced by wind and live load.

7 上一页下一页回目录 4.1 Introduction If a truss carries equal or nearly equal loads at all panel points, the direction in which the diagonals slope will determine if they carry tension or compression forces. for example, the figure 4.3 shows the difference in forces set up in the diagonals of two trusses that are identical in all respects (same span, same loads, and so forth) except for the direction in which the diagonals slope

8 上一页下一页回目录 4.1 Introduction Although trusses are very stiff in their own plane, they are very flexible out of plane and must be braced or stiffened for stability. connect several trusses together to form a rigid-box type of structure.

9 上一页下一页回目录 4.2 Types of Trusses The members of most modern trusses are arranged in triangular patterns because even when the joints are pinned, the triangular form is geometrically stable and will not collapse under load A pin-connected rectangular element, which acts like an unstable linkage, will collapse under the smallest lateral load.

10 上一页下一页回目录 4.2 Types of Trusses One method to establish a stable truss is to construct a basic triangular unit (the shaded triangular element ABC) and then establish additional joints by extending bars from the joints of the first triangular element We can imagine that joint E is formed by extending bars from joints C and D. simple trusses.

11 上一页下一页回目录 4.2 Types of Trusses If two or more simple trusses are connected by a pin or a pin and a tie, the resulting truss is termed a compound truss If a truss — usually one with an unusual shape — is neither a simple nor a compound truss, it is termed a complex truss

12 上一页下一页回目录 4.3 Analysis of Trusses To compute the reactions of a determinate truss, we treat the entire structure as a rigid body and, apply the equations of static equilibrium together with any condition equations that may exist. Three assumptions: 1. Bars are straight and carry only axial load ; 2. Members are connected to joints by frictionless pins; 3. Loads are applied only at joints.

13 上一页下一页回目录 4.3 Analysis of Trusses We label a tensile force positive and a compression force negative. We denote the sense of a force by adding after its numerical Value a T to indicate a tension force and a C to indicate a compression force. If a bar is in tension, the axial forces at the ends of the bar act outward and tend to elongate the bar. If a bar is in compression, the axial forces at the ends of the bar act inward and compress the bar

14 上一页下一页回目录 4.4 Method of Joints To determine bar forces by the method of joints, we analyze free-body diagrams of joints to determine the bar forces in members AB and BC, we use the free body of joint B shown in Figure only two equations are available:

15 上一页下一页回目录 4.4 Method of Joints To determine bar forces by writing out the equilibrium equations, we must assume a direction for each unknown bar force. The analyst is free to assume either tension or compression for any unknown bar force (we can assume that all bars are in tension). The forces are resolved into their X and Y (rectangular) components. Write and solve the two equations of equilibrium.

16 上一页下一页回目录 4.4 Method of Joints If the solution of an equilibrium equation produces a positive value of force, the direction initially assumed for the force was correct. If the value of force is negative, its magnitude is correct, but the direction initially assumed is incorrect, and the direction of the force must be reversed on the sketch of the free-body diagram. After the bar forces are established at a joint, the engineer proceeds to adjacent joints and repeats the preceding computation until all bar forces are evaluated.

17 上一页下一页回目录 4.4 Method of Joints Determination of Bar Forces by Inspection Trusses can often be analyzed rapidly by inspection of the bar forces and loads acting on a joint that contains one sloping bar in which the force is unknown. For example To determine the force F BC, we mentally sum forces in the x direction. Since the applied load of 30 kips at joint B is directed downward, Y AB must be equal to 30 kips and directed upward, F AB must act upward and to the right, X AB must be directed to the right. The value of X AB is easily computed from similar triangles.

18 上一页下一页回目录 4.4 Method of Joints E X A M P L E 4. 1 Analyze the truss by the method of joints. Reactions are given.

19 上一页下一页回目录 4.4 Method of Joints Solution Since the computations are simplest at a joint with one sloping member, we start at A. On a free body of joint A, we arbitrarily assume that bar forces F AB and F AD are tensile forces and show them acting outward on the joint. We next replace F AB by its rectangular components X AB and Y AB. Since Y AB is positive, it is a tensile force, and the assumed direction on the sketch is correct.

20 上一页下一页回目录 4.4 Method of Joints Solution Compute X AB and F AB by proportion, considering the slope of the bar. Compute F AD Since the minus sign indicates that the direction of force F AD was assumed incorrectly, the force in member AD is compression

21 上一页下一页回目录 4.4 Method of Joints Solution Isolate joint B and show all forces acting on the joint Since Y BD =0, it follows that F BD =0 Compute F BC :

22 上一页下一页回目录 4.4 Method of Joints Solution Analyze joint D with F BD =0 and F DC shown as a compressive force As a check of the results, we observe that the components of F DC are proportional to the slope of the bar. Since all bar forces are known at this point, we can also verify that joint C is in equilibrium, as an alternative check.

23 上一页下一页回目录 4.5 Zero Bars Case 1:If No External Load Is Applied to a Joint That Consists of Two Bars, the Force in Both Bars Must Be Zero The joint cannot be in equilibrium unless Y 2 equals zero because no other force is available to balance Y 2. If Y 2 equals zero, then F 2 is zero, and equilibrium requires that F 1 also equal zero.

24 上一页下一页回目录 Case 2: If No External Load Acts at a Joint Composed of Three Bars — Two of Which Are Collinear — the Force in the Bar That Is Not Collinear Is Zero The equilibrium equation can be satisfied only if F 3 equals zero because there is no other force to balance its y-component Y Zero Bars

25 上一页下一页回目录 E X A M P L E Zero Bars Based on the earlier discussion,label all the bars in the truss that are unstressed when the 60-kip load acts.

26 上一页下一页回目录 4.5 Zero Bars Solution Although the two cases discussed in this section apply to many of the bars, we will examine only joints A, E, I, and H. Since joints A and E are composed of only two bars and no external load acts on the joints, the forces in the bars are zero

27 上一页下一页回目录 4.5 Zero Bars The horizontal reaction at I is zero. At joint I the force in bar IJ and the 180-kip reaction are collinear The force in bar IH must equal zero. A similar condition exists at joint H. the component of bar HJ must be zero. Solution

28 上一页下一页回目录 4.6 Method of Sections To analyze a stable truss by the method of sections, we imagine that the truss is divided into two free bodies by passing an imaginary cutting plane through the structure. The cutting plane must, pass through the bar whose force is to be determined. At each point where a bar is cut, the internal force in the bar is applied to the face of the cut as an external load. We often use sections that cut three bars since three equations of static equilibrium are available to analyze a free body.

29 上一页下一页回目录 4.6 Method of Sections If we wish to determine the bar forces in the chords and diagonal of an interior panel of the truss in the right figure. We can pass a vertical section through the truss, producing the free-body diagram. The engineer is free to assume the direction of the bar force.

30 上一页下一页回目录 4.6 Method of Sections E X A M P L E 4. 3 Using the method of sections, compute the forces or components of force in bars HC, HG, and BC of the truss

31 上一页下一页回目录 4.6 Method of Sections Solution Pass section 1-1 through the truss cutting the free body. To simplify the computations, force F HC is resolved into vertical and horizontal components. Compute Y HC

32 上一页下一页回目录 4.6 Method of Sections From the slope relationship Compute F BC. Sum moments about an axis through H at the intersection of forces F HG and F HC. Solution

33 上一页下一页回目录 4.6 Method of Sections Compute F HG Since the solution of the equilibrium equations above produced positive values of force, the directions of the forces shown in Figure are correct. Solution

34 上一页下一页回目录 4.6 Method of Sections E X A M P L E 4. 4 Analyze the determinate truss in Figure to determine all bar forces and reactions.

35 上一页下一页回目录 4.6 Method of Sections Solution Since the supports at A, C, and D supply four restraints to the truss and only three equations of equilibrium are available, we cannot determine the value of all the reactions by applying the three equations of static equilibrium Since only one horizontal restraint exists at support A, we can determine its value by summing forces in the x-direction.

36 上一页下一页回目录 4.6 Method of Sections Solution Pass a vertical section through the center panel of the truss. Compute A y Compute F BC

37 上一页下一页回目录 4.6 Method of Sections Solution Compute F FE Now that several internal bar forces are known, we can complete the analysis using the method of joints. Isolate joint E

38 上一页下一页回目录 4.6 Method of Sections Solution Since the slope of bar ED is 1:1, Y ED =X ED =80 kips. The balance of the bar forces and the reactions at C and D can be determined by the method of joints. Final results are shown on a sketch of the truss in the right figure.

39 上一页下一页回目录 4.6 Method of Sections E X A M P L E 4. 5 Determine the forces in bars HG and HC of the truss by the method of sections.

40 上一页下一页回目录 4.6 Method of Sections Solution Pass vertical section 1-1 through the truss, and consider the free body to the left of the section The bar forces are applied as external loads to the ends of the bars at the cut. Since three equations of statics are available, all bar forces can be determined by the equations of statics. To simplify the computations, we select a moment center (point a that lies at the intersection of the lines of action of forces F 1 and F 3 ).

41 上一页下一页回目录 4.6 Method of Sections Solution Force F 2 is next extended along its line of action to point C and replaced by its rectangular components X 2 and Y 2. The distance x between a and the left support is established by proportion using similar triangles, that is, aHB and the slope (1: 4) of force F 1. Sum moments of the forces about point a and solve for Y 2.

42 上一页下一页回目录 4.6 Method of Sections Solution Based on the slope of bar HC, establish X 2 by proportion. Now compute the force F 1 in bar HG. Select a moment center at the intersection of the lines of action of forces F 2 and F 3, that is, at point C

43 上一页下一页回目录 4.6 Method of Sections Solution Establish Y 1 by proportion. Extend force F 1 to point G and break into rectangular components. Sum moments about point C.

44 上一页下一页回目录 4.6 Method of Sections E X A M P L E 4. 6 Using the method of sections, compute the forces in bars BC and JC of the K truss

45 上一页下一页回目录 4.6 Method of Sections Solution To compute the force in bar BC, we pass section 1-1 through the truss The free body to the left of the section is shown in the figure Summing moments about the bottom joint G gives

46 上一页下一页回目录 4.6 Method of Sections Solution To compute F JC, we pass section 2-2 through the panel and consider again the free body to the left The K truss can also be analyzed by the method of joints by starting from an outside joint such as A or H.

47 上一页下一页回目录 4.7 Determinacy and Stability Since indeterminate trusses are also used in practice, an engineer must be able to recognize a structure of this type because indeterminate trusses require a special type of analysis. If you are responsible for establishing the configuration of a truss for a special situation, you must obviously be able to select an arrangement of bars that is stable. If load is applied only at the joints and if all truss members are assumed to carry only axial load, the forces acting on a free-body diagram of a joint will constitute a concurrent force system.

48 上一页下一页回目录 To be in equilibrium, a concurrent force system must satisfy the following two equilibrium equations: Since we can write two equilibrium equations for each joint in a truss, the total number of equilibrium equations available to solve for the unknown bar forces b and reactions r equals 2n If a truss is stable and determinate, the relationship between bars, reactions, and joints must satisfy the following criteria: 4.7 Determinacy and Stability

49 上一页下一页回目录 Although three equations of statics are available to compute the reactions of a determinate truss, these equations are not independent and they cannot be added to the 2n joint equations. If the resultant is zero, the equations of static equilibrium are automatically satisfied when applied to the entire structure and thus do not supply additional independent equilibrium equations. If The number of unknown forces exceed the available equations of statics and the truss is indeterminate. The degree of indeterminacy D equals: 4.7 Determinacy and Stability

50 上一页下一页回目录 If There are insufficient bar forces and reactions to satisfy the equations of equilibrium, and the structure is unstable. If you are uncertain about the stability of a structure, analyze the structure for any arbitrary loading. If a solution that satisfies statics results, the structure is stable. The analysis of an unstable structure leads to an inconsistent equilibrium equation. 4.7 Determinacy and Stability

51 上一页下一页回目录 Since b+r=2n and the reactions are not equivalent to either a concurrent or a parallel force system, the truss is stable and determinate. Since b+r exceeds 2n (18>16), the structure is indeterminate to the second degree Because b+r=2n, and the supports are not equivalent to either a parallel or a concurrent force system, the structure appears stable. We can confirm this conclusion by observing that truss ABC is obviously a stable component of the structure, Since the hinge at C is attached to the stable truss on the left, it is a stable point in space. we can reason that truss CD must also be stable 4.7 Determinacy and Stability

52 上一页下一页回目录 Treat triangular element BCE as a three-bar truss (b=3) supported by three links — AB, EF, and CD (r=3). Since the truss has three joints (B, C, and E), n=3. And b+r=6 equals 2n=2(3) =6, the structure is determinate and stable. b+r=2n, it appears the structure is stable and determinate; since a rectangular panel exists between joints B, C, G, and H, we will verify that the structure is stable by analyzing the truss for an arbitrary load of 1 kips applied vertically at joint D. 4.7 Determinacy and Stability

53 上一页下一页回目录 The center panel, lacking a diagonal bar, cannot transmit vertical force. The structure is unstable. B=16 r=4 n=10 Although b+r=2n, the small truss on the right (DEFG) is unstable. Truss is geometrically unstable. 4.7 Determinacy and Stability

54 上一页下一页回目录 B=21 r=3 n=10 Truss is indeterminate to the fourth degree. B=6 r=3 n=5 Because there are fewer restraints than required by the equations of statics, the structure is unstable. B=9 r=3 n=6 Because the small triangular truss ABC at the top is supported by three parallel links, the structure is unstable. 4.7 Determinacy and Stability

55 上一页下一页回目录 E X A M P L E 4. 7 Verify that the truss is stable and determinate by demonstrating that it can be completely analyzed by the equations of statics for a force of 4 kips at joint F. Solution We must analyze it by the method of joints. We first determine the zero bars. Next we analyze in sequence joints F, C, G,H, A, and B. Since all bar forces and reactions can be determined by the equations of statics, we conclude that the truss is stable and determinate. 4.7 Determinacy and Stability

56 上一页下一页回目录 E X A M P L E 4. 8 Prove that the truss is unstable by demonstrating that its analysis for a load of arbitrary magnitude leads to an inconsistent equation of equilibrium. 4.7 Determinacy and Stability

57 上一页下一页回目录 Solution Apply a load at joint B, say 3 kips, and compute the reactions, considering the entire structure as a free body. Equilibrium of joint B requires that F BF =3 kips tension. Equilibrium in the x direction is possible if F AB = F BC. 4.7 Determinacy and Stability

58 上一页下一页回目录 consider joint F, F AF = 3 kips, compression, X FE = 3 kips, X AF = 3 kips examine support A Since the equilibrium equation is not satisfied, the structure is not stable. 4.7 Determinacy and Stability

59 上一页下一页回目录 4.8 Computer Analysis of Trusses The preceding sections of this chapter have covered the analysis of trusses based on the assumptions that (1) members are connected at joints by frictionless pins. (2) loads are applied at joints only. To analyze a truss with rigid joints would be a lengthy computation by the classical methods of analysis. Now that computer programs are available, we can analyze both determinate and indeterminate trusses as a rigid-jointed structure to provide a more precise analysis, and the limitation that loads must be applied at joints is no longer a restriction.

60 上一页下一页回目录 4.8 Computer Analysis of Trusses Because computer programs require values of cross-sectional properties of members — area and moment of inertia — members must be initially sized. The assumption of pin joints will permit you to compute axial forces that can be used to select the initial cross-sectional areas of members. To carry out the computer analyses, we will use the RISA-2D computer program that is located on the website of

61 上一页下一页回目录 Summary Trusses are composed of slender bars that are assumed to carry only axial force. Although trusses are stiff in their own plane, they have little lateral stiffness; therefore, they must be braced against lateral displacement at all panel points. To be stable and determinate, the following relationship must exist among the number of bars b, reactions r, and joints n: b+r=2n If b+r<2n, the truss is unstable. If b+r>2n, the truss is indeterminate.

62 上一页下一页回目录 Summary Determinate trusses can be analyzed either by the method of joints or by the method of sections. If the analysis of a truss results in an inconsistent value of forces, that is, one or more joints are not in equilibrium, then the truss is unstable.

63 上一页下一页回目录 总结 桁架是由假定只受轴向力的细杆构成的。大型桁架的节点是 将杆件焊接或用螺栓连接到节点板上。如果杆件相对较小或 是承受的应力较小，则节点的通常情形是将竖直或对角杆件 的端点直接焊接到桁架的上弦或下弦。 虽然桁架在其自身平面内刚度很大，但它们的侧向刚度很小； 因此在所有的结点处，它们都必须添加支撑以抵抗侧向位移。 为达到几何不变且静定，以下关于杆件数 b 、反力数 r 和节点 数 n 的关系式必须成立： 另外，反力产生的约束不能组成平行力系或汇交力系。如果， 则桁架为几何可变。如果， 则桁架为超静定。

64 上一页下一页回目录 静定桁架用节点法或截面法都可以分析。当求一根或两根杆 的轴力时可用截面法。当所有杆件的轴力都要求时，要用节 点法。 如果对一个桁架分析后得出的几个力自相矛盾，即一个或一 个以上的节点不能平衡，则该桁架是几何可变的。 总结