1 George Washington Bridge 本章简要介绍索的特性和分析方法，主要包括索的特征、 索内力的变化、重力（竖向荷载作用下索的分析、广义索定 理以及确定索状拱的方法。与桁架相比，索只能承受轴向拉 力，内力单一，但必须形成一定的曲线才能承受不与索轴重 合的荷载，使其构成和分析独具特色。将索变形限定在较小 范围时，通过与相应简支梁对比可得到索分析的一般公式。 本章内容是分析大跨建筑和各种利用预拉力的新型结构的基 础，具有重要作用。学习本章仍需熟练应用平衡方程，并需 加强与梁分析的类比。 CHAPTER 6 Cables

2 上一页下一页回目录 chapter 6 Cables 6.1 Introduction 6.2 Characteristics of Cables 6.3 Variation of Cable Force 6.4 Analysis of a Cable Supporting Gravity (Vertical) Loads 6.5 General Cable Theorem 6.6 Establishing the Funicular Shape of an Arch Summary

3 上一页下一页回目录 6.1 Introduction To use cable construction effectively, the designer must deal with two problems: 1. Preventing large displacements and oscillations from developing in cables that carry live loads whose magnitude or direction changes with time. 2. Providing an efficient means of anchoring the large tensile force carried by cables. Because of their great strength-to-weight ratio, designers use cables to construct long-span structures, including suspension bridges and roofs over large arenas and convention halls.

4 上一页下一页回目录 6.1 Introduction To take advantage of the cable ’ s high strength while minimizing its negative features, designers must use greater inventiveness and imagination than are required in conventional beam and column structures. The small center ring, loaded symmetrically by the cable reactions, is stressed primarily in direct tension while the outer ring carries mostly axial compression.

5 上一页下一页回目录 6.1 Introduction By creating a self-balancing system composed of members in direct stress, the designer creates an efficient structural form for gravity loads that requires only vertical supports around its perimeter. In a typical cable analysis the designer establishes the position of the end supports, the magnitude of the applied loads, and the elevation of one other point on the cable axis The designer applies cable theory to compute the end reactions, the force in the cable at all other points, and the position of other points along the cable axis.

6 上一页下一页回目录 6.2 Characteristics of Cables Cables have an ultimate tensile strength of approximately 270 kips/in 2 (1862 MPa). While the drawing of wires through dies during the manufacturing process raises the yield point of the steel, it also reduces its ductility. Wires can undergo an ultimate elongation of 7 or 8 percent with a moderate yield point, say, 36 kips/in 2 (248 MPa). Steel cables have a modulus of elasticity of approximately 26,000 kips/in 2 (179 GPa) compared to a modulus of 29,000 kips/in 2 (200 GPa) for structural steel bars.

7 上一页下一页回目录 6.2 Characteristics of Cables Since a cable carries only direct stress, the resultant axial force T on all sections must act tangentially to the longitudinal axis of the cable Designers must use great care when designing cable structures to ensure that live loads do not induce either large deflections or vibrations. The complete destruction of the Tacoma Narrows Bridge on November 7, 1940, by wind-induced oscillations is one of the most spectacular examples of a structural failure of a large cable-supported structure.

8 上一页下一页回目录 6.3 Variation of Cable Force If a cable supports vertical load only, the horizontal component H of the cable tension T is constant at all sections along the axis of the cable. If the cable tension is expressed in terms of the horizontal component H and the cable slope θ The maximum value of T typically occurs at the support where the cable slope is largest.

9 上一页下一页回目录 6.4 Analysis of a Cable Supporting Gravity (Vertical) Loads When a set of concentrated loads is applied to a cable of negligible weight, the cable deflects into a series of linear segments The resulting shape is called the funicular polygon. The forces acting at point B on a cable segment of infinitesimal length. the vector diagram consisting of the cable forces and the applied load forms a closed force polygon

10 上一页下一页回目录 6.4 Analysis of a Cable Supporting Gravity (Vertical) Loads The vector diagram consisting of the cable forces and the applied load forms a closed force polygon A cable supporting vertical load is a determinate member. Since the moment at all sections of the cable is zero, the condition equation can be written at any section as long as the cable sag is known.

11 上一页下一页回目录 6.4 Analysis of a Cable Supporting Gravity (Vertical) Loads The cable sag at the location of the 12-kip load is set at 6 ft. we will assume that the weight of the cable is trivial and neglect it. STEP 1 Compute D y by summing moments about support A. STEP 2 Compute A y.

12 上一页下一页回目录 6.4 Analysis of a Cable Supporting Gravity (Vertical) Loads STEP 3 Compute H; sum moments about B After H is computed, we can establish the cable sag at C by considering a free body of the cable just to the right of C

13 上一页下一页回目录 6.4 Analysis of a Cable Supporting Gravity (Vertical) Loads STEP 4 To compute the force in the three cable segments, we establish θ A, θ B, and θ C and then use Equation 6.1.

14 上一页下一页回目录 6.5 General Cable Theorem You may have observed that certain of the computations are similar to those you would make in analyzing a simply supported beam with a span equal to that of the cable and carrying the same loads applied to the cable. We apply the cable loads to a beam whose span equals that of the cable. If we sum moments about support A to compute the vertical reaction D y at the right support, the moment equation is identical to Equation 6.2 previously written to compute the vertical reaction at the right support of the cable.

15 上一页下一页回目录 6.5 General Cable Theorem At any point on a cable supporting vertical loads, the product of the cable sag h and the horizontal component H of the cable tension equals the bending moment at the same point in a simply supported beam that carries the same loads in the same position as those on the cable. The span of the beam is equal to that of the cable. the general cable theorem: The relationship above can be stated by the following equation:

16 上一页下一页回目录 6.5 General Cable Theorem H: horizontal component of cable tension h z : cable sag at point z where Mz is evaluated Since H is constant at all sections, Equation 6.6 shows that the cable sag h is proportional to the ordinates of the moment curve. M z :moment at point z in a simply supported beam carrying the loads applied to the cable

17 上一页下一页回目录 6.5 General Cable Theorem We will show that at an arbitrary point z on the cable axis the product of the horizontal component H of cable thrust and the cable sag h z equals the moment at the same point in a simply supported beam carrying the cable loads The vertical distance between the two supports can be expressed as

18 上一页下一页回目录 6.5 General Cable Theorem Directly below the cable we show a simply supported beam to which we apply the cable loads. The arbitrary section at which we will evaluate the terms in Equation 6.6 is located a distance x to the right of the left support. We begin by expressing the vertical reaction of the cable at support A in terms of the vertical loads and H :represents the moment about support B of the vertical loads (P 1 through P 4 ) applied to the cable.

19 上一页下一页回目录 6.5 General Cable Theorem Considering a free body to the left of point z, we sum moments about point z represents the moment about z of the loads on a free body of the cable to the left of point z. Solving Equation 6.8 for A y gives

20 上一页下一页回目录 6.5 General Cable Theorem evaluate M z we sum moments of the forces about the roller at B Substituting R A from Equation 6.13 into Equation 6.12 gives Since the right sides of Equations 6.11 and 6.14 are identical, we can equate the left sides, giving, Hh z =M z, and Equation 6.6 is verified.

21 上一页下一页回目录 6.6 Establishing the Funicular Shape of an Arch For a particular set of loads the arch profile in direct stress is called the funicular arch. If the cable shape is turned upside down, the designer produces a funicular arch. Since dead loads are usually much greater than the live loads, a designer might use them to establish the funicular shape

22 上一页下一页回目录 E X A M P L E 6. 1 Determine the reactions at the supports produced by the 120- kip load at midspan. a) using the equations of static equilibrium and (b) using the general cable theorem. Neglect the weight of the cable. Solution 6.6 Establishing the Funicular Shape of an Arch a) consider Figure 6.7a

23 上一页下一页回目录 consider Figure 6.7b. Solution Substitute H from Equation 2 into Equation 1. Substituting C y into Equation 2 yields 6.6 Establishing the Funicular Shape of an Arch

24 上一页下一页回目录 apply Equation 6.6 at midspan where the cable sag h z =8 ft and M z =3000 kipft After H is evaluated, sum moments about A in Figure 6.7a to compute C y = kips. Solution 6.6 Establishing the Funicular Shape of an Arch b)

25 上一页下一页回目录 E X A M P L E 6. 2 A cable-supported roof carries a uniform load w=0.6 kip/ft If the cable sag at midspan is set at 10 ft, what is the maximum tension in the cable (a) between points B and D and (b) between points A and B? 6.6 Establishing the Funicular Shape of an Arch

26 上一页下一页回目录 Apply the uniform load to a simply supported beam and compute the moment M z at midspan The maximum cable tension in span BD occurs at the supports where the slope is maximum. Solution 6.6 Establishing the Funicular Shape of an Arch

27 上一页下一页回目录 Solution To establish the slope at the supports, we differentiate the equation of the cable y=4hx 2 /L 2 at x=60 ft, tanθ=8(10)(60)/(120) 2 =1/3, and θ=18.43°, cosθ= Substituting into b.If we neglect the weight of the cable between points A and B, the cable can be treated as a straight member. Since the cable slopeθ is 45, the cable tension equals. 6.6 Establishing the Funicular Shape of an Arch

28 上一页下一页回目录 Summary ♫ Cables, composed of multiple strands of cold-drawn, high- strength steel wires twisted together, have tensile strengths varying from 250 to 270 ksi. Cables are used to construct long- span structures. ♫ Since cables are flexible, they can undergo large changes in geometry under moving loads; therefore, designers must provide stabilizing elements to prevent excessive deformations. Also the supports at the ends of cables must be capable of anchoring large forces. ♫ Because cables (due to their flexibility) have no bending stiffness,the moment is zero at all sections along the cable.

29 上一页下一页回目录 Summary ♫ The general cable theorem establishes a simple equation to relate the horizontal thrust H and the cable sag h to the moment that develops in a fictitious, simply supported beam with the same span as the cable ♫ When cables are used in suspension bridges, floor systems must be very stiff to distribute the concentrated wheel loads of trucks to multiple suspenders, thereby minimizing deflections of the roadway. ♫ Since a cable is in direct stress under a given loading, the cable shape can be used to generate the funicular shape of an arch by turning it upside down.

30 上一页下一页回目录 总结 ♫ 由多股冷拉高强钢丝拧成的索的抗拉强度在 250 到 270 kips 之间。索通常用来建造大跨结构。 ♫ 因为索很柔软，在移动荷载作用下索能产生很大的几何变形； 所以设计师必须提供稳定构件避免索发生过度变形。索端的支 座必须承受很大的反力。若没有岩床锚固悬索桥的索端，则需 要大型的钢筋混凝土墩座。 ♫ 因为索不具有抗弯刚度（柔软性），所以沿着索的每一个截 面上的弯矩都为零。

31 上一页下一页回目录 ♫ 广义索定理可通过建立简单的等式来确定水平拉力 H 和索的 垂度 h 与假想的等跨简支梁上产生的弯矩的联系： Hh Z ＝ M Z H —— 索拉力的水平分量； h Z —— 计算 M Z 的 z 点处垂度。垂度是索弦与索的竖向距离； M Z —— 等跨且承受相同荷载的简支梁在 z 点处的弯矩。 ♫ 当索用于悬索桥时，桥面要有很大的刚度将汽车或卡车轮子 的集中荷载传递到多个吊杆，使路面的变形最小。 ♫ 在荷载作用下，索处于直接应力状态，将索的形状向上翻转， 即得到索状拱的外形。 总结