Structural Theory Chapter 4 PDF

STRUCTURAL THEORY INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS INTERNAL LOADINGS AT A SPECIFIED POINT As discussed in equations of equilibrium, the internal load at a specified point in a member can be determined by using the method of sections. In general, this loading for a coplanar s...

STRUCTURAL THEORY INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS INTERNAL LOADINGS AT A SPECIFIED POINT As discussed in equations of equilibrium, the internal load at a specified point in a member can be determined by using the method of sections. In general, this loading for a coplanar structure will consist of a normal force N, shear force V, and bending moment M. It should be realized, however, that these loadings actually represent the resultants of the stress distribution acting over the member’s cross-sectional area at the cut section. Once the resultant internal loadings are known, the magnitude of the stress can be determined provided an assumed distribution of stress over the cross-sectional area is specified. SIGN CONVENTION Before presenting a method for finding the internal normal force, shear force, and bending moment, we will need to establih a sign convention to define their “positive” and “negative” values. Although the choice is arbitrary, the sign convention to be adopted here has been widely accepted in structural engineering practice, and is illustrated in the figure. On the left-hand face of the cut member the normal force N acts to the right, the internal shear force V acts downward, and the moment M acts counterclockwise. In accordance with Newton’s third law, an equal but opposite normal force, shear force, and bending moment must act on the right-hand face of the member at the section. Perhaps an easy way to remember this sign convention is to isolate a small segment of the member and note that positive normal force tends to elongate the segment, postive shear tends to rotate the segment clockwise, and positive moment tends to bend the segment concave upward. equation of equilibrium SHEAR AND MOMENT FUNCTIONS The design of a beam requires a detailed knowledge of the variations of the internal shear force V and moment M acting at each point along the axis of the beam. The internal normal force is generally not considered for two reasons: (1) in most cases the loads applied to a beam act perpendicular to the beam’s axis and hence produce only an internak shear force and bending moment, and (2) for design purposes the beam’s resistance to shear, and particularly to bending, is more important than its ability to resist normal force. An inportant exception to this occurs, however, when beams are subjected to compressive axial forces, since the bckling or instability that may occur has to be investiagated. The variations of V and M as a function of the position x of an arbitrary point along the beam’s axis can be obtained by using the method of sections discussed in internal loadings at a specified point. Here, however, it is necessary to locate the imaginary section or cut at an arbitrary distance x from one end of he beam rather than a specific point. In general, the internal shear and moment functions will be discontinuous, or their slope will be discontinuous, at points whre the type or magnitude of the distributed load changes or where concentrated forces or couple moments are applied. Because of this, shear and moment functions must be determined for each region of the beam located between any two discontinuities of loading. For example, coordinates 𝑥1, 𝑥2, and 𝑥3 will have to be used to describe the variation of V and M throughout the length of the beam in the figure. These coordinates will be valid only within regions from A to B for 𝑥1, from B to C for 𝑥2, and from C to D for 𝑥3. Although each of these coordinates has the same origin, as noted here, this does not have to be the case. Indeed, it may be easier to develop the shear and moment functions using coordinates 𝑥1, 𝑥2, 𝑥3 having origins at A, B, and D as shown in the next figure. Here 𝑥1 and 𝑥2 are positive to the riht and 𝑥3 is positive to the left. Shear and moment diagrams for a beam. SHEAR AND MOMENT DIAGRAMS FOR A BEAM If the variations of V and M as function of x obtained in shear and moment function are plotted, the graphs are termed the shear diagram and moment diagram, respectively. In cases where a beam is subjected to several concentrated forces, couples, and distributed loads, plotting V and M versus x can become quite tedious since several functions must be plotted. In this section a simpler method for constructing diagrams is discussed – a method based on differential relations that exist between the load, shear, and moment. To derive these relations, consider the beam AD in the figure, which is subjected to an arbitrary distributed loading w = w(x) and a series of concentrated forces and couples. In the following discussion, the distributed load will be considered positive when the loading acts upward as shown. We will consider the free-body diagram for a small segment of the beam having a length ∆𝑥 in the next figure. Since this segment has been chosen at point x along the beam that is not subjected to a concentrated force or couple, any results obtained will not apply at points of concentrated loading. The internal shear force and bending moment shown on the free-body diagram are assumed to act in the positive direction according to the established sign convention. Note that both the shear force and moment acting on the right face must be increased by a small, finite amount in order to keep the segment in equilibrium. The distributed loading has been replaced by a concentrated force 𝑤(∆𝑥) that acts at fractional distance ∈ (∆𝑥) from the right end, where 0 < ∈ < 1. 1 (For example, if w is uniform or constant, then 𝑤(∆𝑥) will act at ∆𝑥, 2 1 so ∈=.) Applying the equations of equilibrium, we have 2 Dividing by ∆𝑥 and taking the limit as ∆𝑥 → 0, these equations become 𝑑𝑉 =𝑤 𝑒𝑞. 1 𝑑𝑥 Slope of Shear Diagram = Intensity of Distributed Load 𝑑𝑀 =𝑉 𝑒𝑞. 2 𝑑𝑥 Slope of the Moment Diagram = Shear As noted, in the first equation that the slope of the shear diagram at a point (dV/dx) is equal to the intensity of the distributed load w at the point. Likewise, on the second equation that the slope of the moment diagram (dM/dx) is equal to the intensity of the shear at the point. Both equations can be “integrated” from one point to another between concentrated forces or couples (such as from B to C in the first figure), in which case ∆𝑉 = ∫ 𝑤𝑑𝑥 𝑒𝑞. 3 Change in Shear = Area under Distributd Loading Diagram and ∆𝑀 = ∫ 𝑉𝑑𝑥 𝑒𝑞. 4 Change in Moment = Area under Shear Diagram As noted, in the third equation states that the change in the shear between any two points on a beam equals the area under the distributed loading diagram between the points. Likewise, in the forth equation states that the change in the moment between the two points equals the area under the shear diagram between the points. If the areas under the load and shear diagrams are easy to compute, both equations provide a method for determining the numerical values of the shear and moment at various points along a beam. From the derivation it should be noted that eq. 1 and eq. 3 cannot be used at points where a concentrated force acts, since these equations do not account for the sudden change in shear at these point. Similarly, because of a discontinuity of moment in eq. 2 and eq. 4 cannot be used at points where a couple moment is applied. In order to account for these two cases, we must consider the free-body diagrams of differential elements of the beam in the figure which are located at concentrated force and couple moments. Examples of these elements are shown in figure a and b respectively. From figure a it is seen that force equilibrium requires the change in shear to be +↑ ∑ 𝐹𝑦 = 0 ; ∆𝑉 = 𝐹 𝑒𝑞. 5 Thus, when F acts upward on the beam, ∆𝑉 is positive, so that the shear diagram shows a “jump” upward. Likewise, if F acts ownward, the jump (∆𝑉 ) is downward. From figure b, letting ∆𝑥 → 0 , moment equilibrium requires the change in moment to be 𝐶𝑜𝑢𝑛𝑡𝑒𝑟𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 + ∑ 𝑀𝑜 = 0 ; ∆𝑀 = 𝑀′ 𝑒𝑞. 6 In this case, if an external couple moment M’ is applied clockwise, ∆𝑀 is positive, so that the moment diagram jumps upward, and when M acts counterclockwise, the jump (∆𝑀) must be downward. Table 4.1 illustrates application of eq. 1,2,5, and 6 to some common loading cases assuming V and M retain positive values. The slope at various points on each curve is indicated. None of these results should be memorized; rather, each should be studied carefully so that one becomes fully aware of how the shear and moment diagrams can be constructed on the basis of knowing the variation of the slope from the load and shear diagrams, respectively. It would be well worth the time and effort to self-test your understanding of these concepts byy covering over the shear and moment diagram columns in the table and then tring to reconstruct these diagrams on the bais of knowing the loading.  Eq. 1 through Eq. 6 In previous discussion Eq. 3 In previous discussion Eq. 4 SHEAR AND MOMENT DIAGRAMS FOR A FRAME Recall that a frame is composed of several connected memers that are either fixed or pin connected at their ends. The design of these structures often requires drawing the shear and moment diagrams for each of the members. To analyze any problem, we can use the procedure for analysis outlined in Shear and Moment diagrams for a beam. This requires first determining the reactions at the frame supports. Then, using the method of sections, we find the axial force, shear force, and moment acting at the ends of each member. Provided all loadings are resolved into components acting parallel and perpendicular to the members’s axis, the shear and moment diagrams for each member can then be drawn as described previously. When drawing the moment diagram, one of two sign conventons is used in practice. In particular, if the frame is made of reinforced concrete, designers often draw the moment diagram positive on the tesion side of the frame. In othe words, if the moment produces tension on the outer surface of the frame, the moment diagram is drawn positive on this side. Since concrete has a low tensile strength, it will then be possible to tell at a glance on which side of the frame the reinforcement steel must be placed. In this module, however, we will use the opposite sign convention and always draw the moment diagram positive on the compression side of the member. This convention follows that usd for beams discussed in internal loadings at a specified point. MOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION Since beams are used primarily to resist bending stress, it is important that the moment diagram accompany the solution fortheir design. In Shear and Moment diagrams for a beam the moment diagram was constructed by first drawing the shear diagram. If we use the principle of superposition, however, each of the loads on the beam can be treated separately and the moment diagram can then be constructed in a series of partsrather than a single and sometimes complicated shape. It will be shown later in this module that this can be particularly advantageous when applying geometric defection methods to determine both the deflection of a beam and the reactions on statically indeterminate beams. Most loadings on beams in structural analysis will be a combination of the loadings shown in the figure.  To understand how to use the method of superposition to construct the moment diagram consider the simply supported beam at the top of figure a. Here the reactions have been calculated and so the force system on the beam produces a zero force and moment resultant. The moment diagram for this case is shown at the top of figure b. Note that this same moment diagram is produced for the cantilevered beam when it is subjected to the same statically equivalent system ofloads as the simply supported beam. Rather than considering all the loads on this beam simultaneously when drawing the moment diagram, we can instead superimpose the results of the loads acting separately on the three cantilevered beams shown in figure a. Thus, if the moment diagram for each cantilevered beam is drawn, in figure b, the superposition of these diagrams yields the resultant moment diagram for the simply supported beam. For example, from each of the separate moment diagrams, the moment at end A is 𝑀𝐴 = −200 − 300 + 500 = 0, as verified by the top moment diagram in Figure b. In some cases it is often easier to construct and use a separate series of statically equivalent moment diagrams for a beam, rather than construct the beam’s more complicated “resultant” moment diagram.  In a similar manner, we can also simplify construction of the “resultant” moment diagram for a beam by using a superpositin of “simply supported” beams. For example, the loading on the beam shown at the top of figure a is equivalent to the beam loadings shown below it. Consequently, the separate moment diagrams for each of these three beams can be used rather than drawing the resultant moment diagram shown in figure b.