STRUCTURAL-THEORY-CHAPTER-3.pdf

STRUCTURAL THEORY ANALYSIS OF STATICALLY DETERMINATE TRUSSES COMMON TYPES OF TRUSSES A truss is a structure composed of slender elements joined together at their end points. The members commonly used in construction consist of wooden struts, me...

STRUCTURAL THEORY ANALYSIS OF STATICALLY DETERMINATE TRUSSES COMMON TYPES OF TRUSSES A truss is a structure composed of slender elements joined together at their end points. The members commonly used in construction consist of wooden struts, metal bars, angles, or channels. The joint connections are usually formed by bolting or welding the ends of the members to a common plate, called a gusset plate, as shown in the figure, or by simply passing a large bolt or pin through each of the members. Planar trusses lie in a single plane and are often used to support roofs and bridges. ROOF TRUSSES Roof trusses are often used as part of an industrial building frame, such as the one shown. Here, the roof load is transmitted to the truss at the joints by means of a series of purlins. The roof truss along with its supporting columns is termed a bent. Ordinarily, roof trusses are supported either by columns of wood, steel, or reinforced concrete, or by masonry walls. To keep the bent rigid, and thereby capable of resisting horizontal wind forces, knee braces are sometimes used at the supporting columns. The space between adjacent bents is called a bay. Bays are economically spaced at about 15 ft (4.6m) for spans around 60 ft (18m) and about 20 ft (6.1m) for spans of 100 ft (30m). Bays are often tied together using diagonal bracing in order to maintain rigidity of the building’s structure. Trusses used to support roofs are selected on the basis of the span, the slope, and the roof material. Some of the more common types of trusses used are shown in the next figure. In particular, the scissors truss can be used for short spans that require overhead clearance. The Howe and Pratt trusses are used for roofs of moderate span, about 60 ft (18m) to 100 ft (30m). if larger spans are required to support the roof, the Fan truss or Fink truss may be used. These trusses may be built with a cambered bottom chord such as that shown in the figure. If a flat roof or nearly flat roof is to be selected, the Warren truss is often used. Also, the Howe and Pratt trusses may be modified for flat roofs. Sawtooth trusses are often used where column spacing is not objectionable and uniform lighting is important. A textile mill would be an example. The bowstring truss is sometimes selected for garages and small airplane hangars; and the arched truss although relatively expensive, can be used for high rises and long spans such as field houses, gymnasiums, and so on. BRIDGE TRUSSES The main structural elements of a typical bridge truss are shown in the figure. Here is is seen that a load on the deck is first transmitted to stringers, then to floor beams, and finally t the joints of the two supporting side trusses. The top and bottom cords of these side trusses are connected by top and bottom lateral bracing, which serves to resist the lateral forces causes by wind and the sideways caused by moving vehicles on the bridge. Additional stability is provided by the portal and sway bracing. As in the case of many long-span trusses, a roller is provided at one end of a bridge truss to allow for thermal expansion. A few typical forms of bridge trusses currently used for single spans are shown in figure at the next page. In particular, the Pratt, Howe, and Warren trusses are normally used for spans up to 200 ft (61m) in length. The most common form is the Warren truss with veticals. For larger spans, the height of the truss must increase to support the greater moment developed in the center of the span. As a result, a truss with a polygonal upper cord, such as the Parker truss isused for some savings in material. The Warren truss with verticals can also be fabricated in this manner for span up to 300 ft (91m). The greatest economy of material is obtained if the diagonals have slope between 45 degrees and 60 degrees with the horizontal. If this rule is maintained, then for spans greater than 300 ft (91m), the depth of the truss must increase and consequently the panels will get longer. This results in a heavy deck system and, to keep the weight of the deck within to lerable limits subdivided trusses have been developed. Typical examples include the Baltimore and subdivided Warren trusses. Finally, the K -truss an also be used in place of a subdivided truss, since it accomplishes the same purpose. ASSUMPTIONS FOR DESIGN To design both the members and the connections of a truss, it is first necessary to determine the force developed in each member when the truss is subjected to a given loading. In this regard, two important assumptions will be made in order to idealize the truss. 1. The members are joined together by smooth pins. In cases where bolted or welded joint conections are used, this assumption is generally satisfactory provided the center lines of the joining members are concurrent at a point as in the figure of gusset plate. It should be realized, howevr, that the actual connections do give some rigidity to the joint and this in turn introduces bending of the connected members when the truss is subjected to a load. The bending stress developed in the members is called secondary stress, whereas the stress in the members of the idealized truss, having pin-connected joints, is called primary stress. A secondary stress analysis of a truss can be performed using a computer, to be discussed at the end of the subject. For some types of truss geometries these stresses may be large. 2. All loadings are applied at the joints. In most situations, such as for bridge and roof trusses, this assumptions is true. Frequently in the force analysis, the weight of the members is neglected, since the force supported by the members is larg in comparison with their weight. If the weight is to be included in the analysis, it is generally satisfactory to apply it as a vertical force, half of its magnitude applied at each end of the member. Because of these two assumptions, each truss member acts as an axial force member, and therefore the forces acting at the ends of the member must be directed along the axis of the member. If the force tends to elongate the member, it is a tensile force (T), whereas if the force tends to shorten the member, it is a compressive force (C). In the actual design of a truss, it is important to state whether the force is tensile or compressive. Most often, compression members must be made thicker than tension members, because of the buckling or sudden instability that may occur in compression members. CLASSIFICATION OF COPLANAR TRUSSES Before beginning the force analysis of a truss, it is important to classify the truss as simple, compound, or complex, and then to be able to specify its determinacy and stability. SIMPLE TRUSS To prevent collapse, the framework of a truss must be rigid. Obviously, the four-bar frame ABCD in the figure will collapse unless a diagonal, such as AC, is added for support. The simplest framework that is rigid or stable is a triangle. Consequently, a simple truss is constructed by starting with a basic triangular element, such as ABC in the next figure, and connecting two members (AD and BD) to form an additional element. Thus it is seen that as each additional element of two members is placed on the truss, the number of joints is increased by one. An example of a simple truss shown in the third figure, where the basic “stable” triangular element is ABC, from which the remainder of the joints. D, E, and F, are established in alphabetical sequence. For this method of construction, however, it is important to realize that simple trusses do not have to consist entirely of triangles. An example is shown in the forth figure, where starting with triangle ABC, bars CD and AD are added to form joint D. Finally, bars BE and DE are added to form joint E. COMPOUND TRUSS A compound truss is formed by connecting two or more simple trusses together. Quite often this type of truss is used to support loads acting over a large span, since it is cheaper to construct a somewhat lighter compound trussthan to use a heavier single simple truss.  There are three ways in which simple trusses are joined together to form a compound truss. The trusses may be connected by a common joint and bar. Anexample is given in the figure a, where the shaded truss ABC is connected to the shaded truss CDE in this manner. The trusses may be joined by three bars, as in the case of the shaded truss ABC is connected to the larger truss DEF, Figure b. And finally, the trusses may be joined where bars of a large simple truss, called the main truss, have been replaced by simple trusses, called secondary trusses. An example is shown in figure c, where dashed members of the main trus ABCDE have been replaced by the secondary shaded trusses. If this truss carried roof loads, the use of the secondary trusses might be more economical, since the dashed members may be subjected to excessive bending, whereas the secondary trusses can better transfer the load. COMPLEX TRUSS A complex truss is one that cannot be clasified as being either simple or compound. The truss in the figure is an example. DETERMINACY For any problem in truss analysis, it should be realized that the total number of unknowns includes the forces in b number of bars of the truss and the total number of external support reactions r. Since the truss members are all straight axial force members lying in the same plane, the force system acting at each joint is coplanar and concurrent. Consequently, rotational or moment distribution is automatically satisied at the joint (or pin), and it is only necessary to satisfy ∑ 𝐹𝑥 = 0 and ∑ 𝐹𝑦 = 0 to ensure translational or force equilibrium. Therefore, only two equations of equilibrium can be written for each joint, and if there are j number of joints, the total number of equations available for solution is 2j. By simply comparing the total number of unknowns (b + r) with the total number of available equilibrium equations, it is therefore possible to specify the determinacy for either a simple, compound, or complex truss. We have b + r = 2j statically determinate b + r > 2j staticaly indeterminate In paricular, the degree of indeterminacy is specified by the difference in the numbers (b + r) – 2j. STABILITY If b + r < 2j, a truss will be unstable, that is, a loading can be applied to the truss that causes it to collapse, since there will be an insufficient number of bars or reacions to constrain all the joints. Also, a truss can be unstable if it is statically determinate or statically indeterminate. In this cse the stability will have to be determined either by inspection or by a force analysis. EXTERNAL STABILITY As stated in the previous discussion, a structure (or truss) is externally unstable if all its reaction are concurrent or parallel. For example, if a horizontal force is applied to the top chord of each of the two trusses in the figure, each truss will be externally unstable, since the support reactions have lines of action that are either concurrent or parallel. INTERNAL STABILITY The internal stability of a truss can often be checked by careful inspection of the arrangement of its members. If it can be determined that each joint is held fixed so that it cannot move in a “rigid body” sense with respect to the other joints, then the truss will be stable. Notice that a simple truss will always be internally stable, since by the nature of its construction it requires starting from basic triangular element and adding successive “rigidelements,” each containing two additional members and a joint. The truss in the figure exeplifies this construction, where, starting with the shaded triangle element ABC, the successive joints D, E, F, G, H have been added. If a truss is constructed so that it does not hold its joints in a fixed position, it will be unstable or have a “critical form.” An obvious example of this is shown in the next figure, where it can be seen that no restraint or fixity is provided between joints C and F or B and E, and so the truss will collapse under a vertical load. To determine the internal stability of a compound truss, it is necessary to identify the way in which the simple trusses are connected together. For example, the compound truss in the third figure is unstable since the inner simple truss ABC is connected to the outer simple truss DEF using three bars, AD, BE, and CF, which are concurrent at a point O. Thus an external load can be applied to joint A, B, or C and cause the truss ABC to rotate slightly. If a truss is identified as complex, it may not be possible to tell by inspection if it is stable. For example, it can be shown by the analysis that will be discussed in complex trusses, that the complex truss shown in the figure is unstable or has a “critical form” only if the dimension d = d’. If d ≠ d’ it is stable. The instability of any form of truss, be it simple, compound, or complex, may also be noticed by using a computer to solve 2j simultaneous equations written for all joints of the truss. If inconsistent results are obtained, the truss will be unstable or have a critical form. If a computer analysis is not performed, the methods discussed previously can be used to check the stability of the truss. To summarize, if the truss has b bars, r external reactions, and j joints, then if b + r < 2j unstable b + r ≥ 2j unstable if truss support reactions are concurrent or parallel or if some of the components of the truss form a collapsible mechanism  Bear in mind, however, that if a truss is unstable, it does not matter whether it is statically determinate or indeterminate. Obviously, the use of an unstable truss is to be avoided in practice. THE METHODS OF JOINTS If a truss is in equilibrium, then each of its joints must also be in equilibrium. Hence, the method of joints consists of satisfying the equilibrium conditions ∑ 𝐹𝑥 = 0 and ∑ 𝐹𝑦 = 0 for the forces exerted on the pin at each joint of the truss. When using the method of joints, it is necessary to draw each joint’s free- body diagram before applying the equilibrium equations. Recall that the line of action of each member force acting on the joint is specified from the geometry of the truss, since the force in a member passes along the axis of the member. As an example, consider joint B of the truss in figure a. From the free-body diagram, figure b, the only unknowns are the magnitudes of the forces in members BA and BC. As shown, 𝐹𝐵𝐴 is “pulling” on the pin, which indicates that member BA is in tension, whereas 𝐹𝐵𝐶 is “pushing” on the pin, and consequently member BC is in cmpression. These effects are clearly demonstrated by drawing the free-body diagrams of the connected members, figure c. In all cases, the joint analysis should start at a joint having atleast one known force and at two most unknown forces, as in figure b. In this way, application of ∑ 𝐹𝑥 = 0 and ∑ 𝐹𝑦 = 0 yields two algebraic equations that can be solved for the two unknowns. When applying these equations, the correct sense of an unknown member force can be determined using one of two possible methods. 1. Always assume the unknown member forces acting on the joint’s free - body diagram to be in tension, i.e., “pulling” on the pin. If this is done, then numerical solution of the equilibrium equations will yield positive scalars for members in tension and negative scalars  for members in compression. Once an unknown member force is found, use its corrects magnitude and sense (T or C) on subsequent joint free-body diagrams. 2. The correct sense of direction of an unknown member force can, in many cases, be determined “by inspection.” For example, 𝐹𝐵𝐶 in figure b must push on the pin (compression) since its horizontal component 𝐹𝐵𝐶 sin45, must balance the 500N force ( ∑ 𝐹𝑥 = 0). Likewise, 𝐹𝐵𝐴 is a tensile force since it balances the vertical component, 𝐹𝐵𝐶 cos45 (∑ 𝐹𝑦 = 0 ). In more complicated cases, the sense of an unknown member force can be assumed; then, after applying the eqilibrium equations, the assumed sense can be verified from the numerical results. A positive answer indicates that the sens is correct, whereas a negative answer indicates that the sense shown on the free-body diagram must be reversed. This is the method we will use in the example problems which follow. ZERO-FORCE MEMBERS Truss analysis using the method of joints is greatly simplified if one is able to first determine those members that no loading. These zero- force members may be necessary for the stability of the truss during construction and to provide support if the applied loading is changed. The zero-force members of a truss can generally be determined by inspection of the joints, and they occur in two cases. CASE 1. (Refer to first figure) Consider the truss in the figure a. The two members at joint C are connected together at a right angle and there is no external load on the joint. The free-body diagram of joint C, figure b, indicates that the force in each member must be zero in order to maintain equilibrium. Furthermore, as in the case of joint A, figure c, this must be true regardless of the angle, say 𝞱, between the members. CASE 2. (Refer to second figure) Zero-force members also occur at joints having a geometry as joint D in the figure a. Here no external load acts on the joint, so that a force summation in the y direction, figure b, which is perpendicular to the two collinear members, requires that 𝐹𝐷𝐹 = 0. Using this result, FC is also a zero-force member, as indicated by the force analysis of joint F, figure c. In summary, then, if only two non-collinear members form a truss joint and no external load or support reaction is applied to the joint, the members must be zero-force members, Case 1. Also, if three members form a truss joint for which two of the members are collinear, the third member is a zero-force member, provided no external force or support reaction is applied to the joint, Case 2. Particular attention should be directed to these conditions of joint geometry and loading, since the analysis of a truss can be considerably simplified by first spotting the zero- force members. THE METHOD OF SECTIONS If the forces in only a few members of a truss are to be found, the method of sections generally provides the most direct means of obtaining these forces. The method of sections consists of passing an imaginary section through the truss, thus cutting it into two parts. Provided the entire truss is in equilibrium, each of the two parts must also be in equilibrium; and as a result, the three equations of equilibrium may be applied to either one of these two parts to determine the member forces at the “cut section.” When the method of sections is used to determine the force in a particular member, a decision must be made as to how to “cut” or section the truss. Since only three independent equilibrium equations ( ∑ 𝐹𝑥 = 0, ∑ 𝐹𝑦 = 0, ∑ 𝑀𝑜 = 0) can be applied to the isolated portion of the truss, try to select a section that, in general, passes through not more that three members in which the forces are unknown. For example, consider the truss in figure a. If the force in member GC is to be determined, section aa will be appropriate. The free-body diagrams of the two parts are shown in figure b and c. In particular, note that the line of action of each force in a sectioned member is specified from the geometry of the truss, since the force in a member passes along the axis of the member. Also, the member forces acting on one part of the truss are equal but opposite to those acting on the other part – Newton’s third law. As shown, members assumed to be in tension ( BC and GC) are subjected to a “pull”, whereas the member in compression (GF) is subjected to a “push.” The three unknown member forces 𝐹𝐵𝐶 , 𝐹𝐺𝐶 , 𝐹𝐺𝐹 can be obtained by applying the three equilibrium equations to the free-body diagram in figure b. If, however, the free-body diagram in figure c is considered, th three support reactions 𝐷𝑥 , 𝐷𝑦 , 𝐸𝑥 will have to be determined first. Why? (This, of course, is done in the usual manner by considering a free -body diagram of the entire truss). When applying the equilibrium equations, consider ways of writing the equations so as yield a direct solution for each of the unknowns, rather than having to solve simultaneous equations. For example, summing moments about C in figure b would yield a direct solution for 𝐹𝐺𝐹 since 𝐹𝐵𝐶 and 𝐹𝐺𝐶 create zero moment abut C. Likewise, 𝐹𝐵𝐶 can be obtained directly by summing moments about G. Finally, 𝐹𝐺𝐶 can be found direcly from a force summation in the vertical direction, since 𝐹𝐺𝐹 and 𝐹𝐵𝐶 have no vertical components. As in the method of joints, there are two ways in which one can determine the correct sense o an unknown member force. 1. Always assume that the unknown member forces at the cut section are in tension, i..e, “pulling” on the member. By doing this, the numerical solution of the equilibrium equations will yield positive scalars for members in tension and negative scalars for members in compression. 2. The correct sense of an unknown member force can in many cases be determined “by inspection”. For example, 𝐹𝐵𝐶 is a tensile force as represented in figure b, since the moment equilibrium G requires that 𝐹𝐵𝐶 create a moment opposite to that of 1000 N force. Also, 𝐹𝐺𝐶 is a tensile force since its vertical component must balance the 1000 N force. In more complicated cases, the sense of an unknown member force may be assumeb. If the solution yields a negative scalar, it indicates tha the force’s sense is opposite to that shown on the free-body diagram. COMPOUND TRUSSES In classification of coplanar trusses it was stated that compound trusses are formed by connecting two o more simple trusses together either by bars or by joins. Occasionally this type of truss I best analyzed by applying both the method of joints and the method of sections. It is often convenient to first recognize the type of construction as listed in classification of coplanar structure and then perform the analysis. The following examples illustrate the procedure. COMPLEX TRUSSES The member forces in a complex truss can be determined using the method of joints; however, the solution will require writing the two equilibrium equations for each of the j joints of the truss and then by solving the complete set of 2j equations simultaneously. This approach may be impractical for hand calculations, especially in the case of large trusses. Therefore, a more direct method for analyzing a complex truss, referred to as method of substitute members, will be presented. SPACE TRUSSES A space truss consists of members joined together at their ends to form a stable three-dimensional structure. In classification of coplanar trusses, it was shown that the simplest form of a table two-dimensional truss consists of the members arranged in the form of a triangle. We then built up the simple plane truss from this basic triangular element by adding two members at a time to form further elements. In a similar manner, the simplest element of a stable space truss is a tetrahdron, formed by connecting six members together with four joints in the figure. Any additional members added to this basic element would be redundant in suporting the force P. A simple space truss can be built from this basic tetraedral element by adding three additional members and another joint forming multiconnected tetrahedrons. DETERMINACY AND STABILITY Realizing that in three dimensions there are three equations of equilibrium available foreach joint ( ∑ 𝐹𝑥 = 0, ∑ 𝐹𝑦 = 0, ∑ 𝐹𝑧 = 0), then for a space truss with j number of joints, 3j equations are available. If the truss has b number of bars and r number of reactions, then like the case of a planar truss we can write b + r < 3j unstable truss b + r = 3j statically determinate – check stability b + r > 3j statically indeterminate – check stability The external stability of the space truss requires that the support reactions keep the truss in force and moment equilibrium along and about any and all axes. This can sometimes be checked by inspection, although if the truss is unstable a solution of the equilibrium equations will give inconsistent results. Internal stability can sometimes be checked by careful inspection of the member arrangement. Provided each joint is held fixed by its supports or connecting members, so that it cannot move with respect to the other joints, the truss can be classified as internally stable. Also, if we do a force analysis of the truss and obtain inconsistent results, then the truss configuration will be unstable or have a “critical form”. ASSUMPTIONS FOR DESIGN The members of a space truss may be treated as axial-force members provided the external loading is applied at the joints and the joints consist of ball-and-socket connections. This assumption is justiied provided the joined members at a connection intesect at a common point and the weight of the members can be neglected. In cases where the weight of a member is to be included in the analysis, it is generally satisfactory to apply it as a vertical force , half of its magnitude applied to each end of the member. For the force analysis the supports of a space truss are generally modeled as a short link, plane roller joint, slotted roller joint, or a ball-and-socket joint. Each of these supports and their reactive force components are shon in table 3.1. X, Y, Z, FORCE COMPOENTS Since the analysis of a space truss is three-dimensional, it will often be necessary to resolve the force F in a member into components acting along the x, y, z axes. For example, in the figure shown, member AB has a length l and known projections x, y, z along the coordinate axes. These projections can be related to the member’s length by the equation l = √𝑥 2 + 𝑦 2 + 𝑧 2 Since the force F acts along the axis of the member, then the components of F can be determined by proportion as follows: 𝑥 𝑦 𝑧 𝐹𝑥 = 𝐹 ( ) 𝐹𝑦 = 𝐹 ( ) 𝐹𝑧 = 𝐹 ( ) 𝑙 𝑙 𝑙 ZERO-FORCE MEMBERS In some cases the joint analysis of a truss can be simplified if one is able to spot the zero-force members by recognizing two common cases of joint geomety. CASE 1. If all but one members connected to a joint lie in the same plane , and provided no external load acts on the joint, then the member not lying in the plane of the other members must be subjected to zeroforce. The proof of this statement is shown in the figure, where members A, B, C lie in the x-y plane. Since the z component of 𝐹𝐷 must be zero to satisfy ∑ 𝐹𝑧 = 0, member D must be a zero-force member. By the same reasoning, member D will carry a load that can be determined from ∑ 𝐹𝑧 = 0 if an external force acts on the joint and has a component acting along the z axis. CASE 2. If it has been determined that all but two of several members connected at a joint support zero force, then e two remaining members must also support zero force, provided they do not lie along the same line. This situation is illustrated in the figure, where it is known that A and C are zero-force members. Since 𝐹𝐷 is collinear with the y axis, then application of ∑ 𝐹𝑥 = 0 or ∑ 𝐹𝑧 = 0 requires the x or z component of 𝐹𝐵 to be zero. Consequently, 𝐹𝐵 = 0. This being the case, 𝐹𝐷 = 0 since ∑ 𝐹𝑦 = 0. Particular attention should be directed to the foregoing two cases of joint geometry and loading, since the analysis of a space truss can be considerably simplified by first spotting the zero-force members.