Civil Engineering app for ANDROID

Civil Engineering app for Android

Most Views

  • Fixed-end beams, continuous beams, continuous trusses, and rigid frames are statically¬†indeterminate. The equations of equilibrium are not sufficient for the deter¬†mination of all the unknown forces and moments. Additional equations based on a¬†knowledge of the deformation of the member are required.

    Hence, while the bending moments in a simply supported beam are determined only by the loads and the span, bending moments in a statically indeterminate member are also a function of the geometry, cross-sectional dimensions, and modulus of elasticity.

    Sign Convention

    For computation of end moments in continuous beams and frames, the following sign convention is most convenient: A moment acting at an end of a member or at a joint is positive if it tends to rotate the joint clockwise, negative if it tends to rotate the joint counterclockwise.
    Similarly, the angular rotation at the end of a member is positive if in a clockwise direction, negative if counterclockwise. Thus, a positive end moment produces a positive end rotation in a simple beam.
    For ease in visualizing the shape of the elastic curve under the action of loads and end moments, bending-moment diagrams should be plotted on the tension side of each member. Hence, if an end moment is represented by a curved arrow, the arrow will point in the direction in which the moment is to be plotted.

    Carry-Over Moments

    When a member of a continuous beam or frame is loaded, bending moments are induced at the ends of the member as well as between the ends. The magnitude of the end moments depends on the magnitude and location of the loads, the geometry of the member, and the amount of restraint offered to end rotation of the member by other members connected to it. Because of the restraint, end moments are induced in the connecting members, in addition to end moments that may be induced by loads on those spans.
    If the far end of a connecting member is restrained by support conditions against rotation, a resisting moment is induced at that end. That moment is called a carryover moment. The ratio of the carry-over moment to the other end moment is called carry-over factor. It is constant for the member, independent of the magnitude and direction of the moments to be carried over. Every beam has two carry-over factors, one directed toward each end.
    As pointed out in Art. 5.10.6, analysis of a continuous span can be simplified by treating it as a simple beam subjected to applied end moments. Thus, it is convenient to express the equations for carry-over factors in terms of the end rotations of simple beams: Convert a continuous member LR to a simple beam with the same span L. Apply a unit moment to one end (Fig. 5.60). The end rotation at the support where the moment is applied is , and at the far end, the rotation is B. By the dummy-load method (Art. 5.10.4), if x is measured from the B end,

    Carry-Over Factors. The preceding equations can be used to determine carryover factors for any magnitude of end restraint. The carry-over factors toward fixed ends, however, are of special importance.
    The bending-moment diagram for a continuous span LR that is not loaded except for a moment M applied at end L is shown in Fig. 5.61a. For determination of the carry-over factor CR toward R, that end is assumed fixed (no rotation can occur there). The carry-over moment to R then is CRM. The moment diagram in Fig. 5.61a can be resolved into two components: a simple beam with M applied at L (Fig. 5.61b) and a simple beam with CRM applied at R (Fig. 5.61c). As indicated in Fig. 5.61d, M causes an angle change at R of . As shown in Fig. 5.61e, CR M induces an angle change at R of CRMR.

    With the use of Eqs. (5.107) and (5.111), the stiffness of a beam with constant moment of inertia is given by

    This equation indicates that a prismatic beam hinged at only one end has threefourths the stiffness, or resistance to end rotation, of a beam fixed at both ends.

    Fixed-End Moments

    A beam so restrained at its ends that no rotation is produced there by the loads is called a fixed-end beam, and the end moments are called fixed-end moments. Fixedend moments may be expressed as the product of a coefficient and WL, where W is the total load on the span L. The coefficient is independent of the properties of other members of the structure. Thus, any member can be isolated from the rest of the structure and its fixed-end moments computed.
    Assume, for example, that the fixed-end moments for the loaded beam in Fig. 5.63a are to be determined. Let MF be the moment at the left end L and MF the L R
    moment at the right end R of the beam. Based on the condition that no rotation is permitted at either end and that the reactions at the supports are in equilibrium with the applied loads, two equations can be written for the end moments in terms of the simple-beam end rotations, L at L and R, at R for the specific loading.
    Let KL be the fixed-end stiffness at L and KR the fixed-end stiffness at R, as given by Eqs. (5.112) and (5.113). Then, by resolution of the moment diagram into simple-beam components, as indicated in Fig. 5.63ƒ to h, and application of the superposition principle (Art. 5.10.6), the fixed-end moments are found to be

    Deflection of Supports. Fixed-end moments for loaded beams when one support is displaced vertically with respect to the other support may be computed with the use of Eqs. (5.116) to (5.121) and the principle of superposition: Compute the fixedend moments induced by the deflection of the beam when not loaded and add them to the fixed-end moments for the loaded condition with immovable supports.
    The fixed-end moments for the unloaded condition can be determined directly from Eqs. (5.116) and (5.117). Consider beam LR in Fig. 5.64, with span L and support R deflected a distance d vertically below its original position. If the beam were simply supported, the angle change caused by the displacement of R would be very nearly d/L. Hence, to obtain the fixed-end moments for the deflected conditions, set @L =@R = d/L and substitute these simple-beam end rotations in Eqs. (5.116) and (5.117):

    where MF is the fixed-end moment at the left support and MF at the right support. L R
    As an example of the use of the curves, find the fixed-end moments in a prismatic beam of 20-ft span carrying a triangular loading of 100 kips, similar to the loading shown in Case 4, Fig. 5.70, distributed over the entire span, with the maximum intensity at the right support.

    Slope-Deflection Equations

    In Arts. 5.11.2 and 5.11.4, moments and displacements in a member of a continuous beam or frame are obtained by addition of their simple-beam components. Similarly, moments and displacements can be determined by superposition of fixed-end-beam components. This method, for example, can be used to derive relationships between end moments and end rotations of a beam known as slope-deflection equations.
    These equations can be used to compute end moments in continuous beams.
    Consider a member LR of a continuous beam or frame (Fig. 5.72). LR may have a moment of inertia that varies along its length. The support R is displaced vertically

    The slope-deflection equations can be used to determine end moments and rotations of the spans of continuous beams by writing compatibility and equilibrium equations for the conditions at each support. For example, the sum of the moments at each support must be zero. Also, because of continuity, the member must rotate through the same angle on both sides of every support. Hence, ML for one span, given by Eq. (5.133) or (5.135), must be equal to MR for the adjoining span, given by Eq. (5.134) or (5.136), and the end rotation  at that support must be the same on both sides of the equation. One such equation with the end rotations at the supports as the unknowns can be written for each support. With the end rotations determined by simultaneous solution of the equations, the end moments can be computed from the slope-deflection equations and the continuous beam can now be treated as statically determinate.
    See also Arts. 5.11.9 and 5.13.2.
    (C. H. Norris et al., ‚Äė‚ÄėElementary Structural Analysis,‚Äô‚Äô 4th ed., McGraw-Hill¬†Book Company, New York.)

    Moment Distribution

    The frame in Fig. 5.74 consists of four prismatic members rigidly connected together
    at O at fixed at ends A, B, C, and D. If an external moment U is applied at  O, the sum of the end moments in each member at O must be equal to U. Furthermore, all members must rotate at O through the same angle , since they are assumed to be rigidly connected there. Hence, by the definition of fixed-end stiffness, the proportion of U induced in the end of each member at O is equal to the ratio of the stiffness of that member to the sum of the stiffnesses of all the members at the joint (Art. 5.11.3).

    can be taken when a member has a hinged end to reduce the work of distributing moments. This is done by using the true stiffness of a member instead of the fixedend stiffness. (For a prismatic beam with one end hinged, the stiffness is threefourth the fixed-end stiffness; for a beam with variable I, it is equal to the fixedend stiffness times 1 - CLCR, where CL and CR are the carry-over factors for the beam.) Naturally, the carry-over factor toward the hinge is zero.
    When a joint is neither fixed nor pinned but is restrained by elastic members connected there, moments can be distributed by a series of converging approximations.
    All joints are locked against rotation. As a result, the loads will create fixed-end moments at the ends of every member. At each joint, a moment equal to the algebraic sum of the fixed-end moments there is required to hold it fixed. Then, one joint is unlocked at a time by applying a moment equal but opposite in sign to the moment that was needed to prevent rotation. The unlocking moment must be distributed to the members at the joint in proportion to their fixed-end stiffnesses and the distributed moments carried over to the far ends.
    After all joints have been released at least once, it generally will be necessary¬†to repeat the process‚ÄĒsometimes several times‚ÄĒbefore the corrections to the fixed ¬†end moments become negligible. To reduce the number of cycles, the unlocking of¬†joints should start with those having the greatest unbalanced moments.

    Suppose the end moments are to be found for the prismatic continuous beam¬†ABCD in Fig. 5.75. The I /L values for all spans are equal; therefore, the relative¬†fixed-end stiffness for all members is unity. However, since A is a hinged end, the¬†computation can be shortened by using the actual relative stiffness, which is 3‚ĀĄ4.
    Relative stiffnesses for all members are shown in the circle on each member. The distribution factors are shown in boxes at each joint.
    The computation starts with determination of fixed-end moments for each member (Art. 5.11.4). These are assumed to have been found and are given on the first line in Fig. 5.75. The greatest unbalanced moment is found from inspection to be at hinged end A; so this joint is unlocked first. Since there are no other members at the joint, the full unlocking moment of +400 is distributed to AB at A and onehalf of this is carried over to B. The unbalance at B now is +400 - 480 plus the carry-over of +200 from A, or a total of -120. Hence, a moment of 120 must be applied and distributed to the members at B by multiplying by the distribution factors in the corresponding boxes.
    The net moment at B could be found now by adding the entries for each member at the joint. However, it generally is more convenient to delay the summation until the last cycle of distribution has been completed.
    The moment distributed to BA need not be carried over to A, because the carryover factor toward the hinged end is zero. However, half the moment distributed to BC is carried over to C.
    Similarly, joint C is unlocked and half the distributed moments carried over to B and D, respectively. Joint D should not be unlocked, since it actually is a fixed end. Thus, the first cycle of moment distribution has been completed.
    The second cycle is carried out in the same manner. Joint B is released, and the distributed moment in BC is carried over to C. Finally, C is unlocked, to complete the cycle. Adding the entries for the end of each member yields the final moments.

    Maximum Moments in Continuous Frames

    In design of continuous frames, one objective is to find the maximum end moments and interior moments produced by the worst combination of loading. For maximum moment at the end of a beam, live load should be placed on that beam and on the  beam adjoining the end for which the moment is to be computed. Spans adjoining these two should be assumed to be carrying only dead load.

    For maximum midspan moments, the beam under consideration should be fully loaded, but adjoining spans should be assumed to be carrying only dead load.

    For maximum midspan moments, the beam under consideration should be fully loaded, but adjoining spans should be assumed to be carrying only dead load.
    The work involved in distributing moments due to dead and live loads in continuous¬†frames in buildings can be greatly simplified by isolating each floor. The¬†tops of the upper columns and the bottoms of the lower columns can be assumed¬†fixed. Furthermore, the computations can be condensed considerably by following¬†the procedure recommended in ‚Äė‚ÄėContinuity in Concrete Building Frames.‚Äô‚Äô
    EB033D, Portland Cement Association, Skokie, IL 60077, and indicated in Fig. 5.74.
    Figure 5.74 presents the complete calculation for maximum end and midspan moments in four floor beams AB, BC, CD, and DE. Building columns are assumed to be fixed at the story above and below. None of the beam or column sections is known to begin with; so as a start, all members will be assumed to have a fixedend stiffness of unity, as indicated on the first line of the calculation.
    On the second line, the distribution factors for each end of the beams are shown, calculated from the stiffnesses (Arts. 5.11.3 and 5.11.4). Column stiffnesses are not shown, because column moments will not be computed until moment distribution to the beams has been completed. Then the sum of the column moments at each joint may be easily computed, since they are the moments needed to make the sum of the end moments at the joint equal to zero. The sum of the column moments at each joint can then be distributed to each column there in proportion to its stiffness.
    In this example, each column will get one-half the sum of the column moments.
    Fixed-end moments at each beam end for dead load are shown on the third line, just above the heavy line, and fixed-end moments for live plus dead load on the fourth line. Corresponding midspan moments for the fixed-end condition also are shown on the fourth line and, like the end moments, will be corrected to yield actual midspan moments.

    Moment-Influence Factors

    In certain types of framing, particularly those in which different types of loading conditions must be investigated, it may be convenient to find maximum end moments from a table of moment-influence factors. This table is made up by listing for the end of each member in the structure the moment induced in that end when a moment (for convenience, +1000) is applied to every joint successively. Once this table has been prepared, no additional moment distribution is necessary for computing the end moments due to any loading condition.
    For a specific loading pattern, the moment at any beam end MAB may be obtained from the moment-influence table by multiplying the entries under AB for the various  joints by the actual unbalanced moments at those joints divided by 1000, and summing (see also Art. 5.11.9 and Table 5.6).

    Procedure for Sidesway

    Computations of moments due to sidesway, or drift, in rigid frames is conveniently executed by the following method:
    1. Apply forces to the structure to prevent sidesway while the fixed-end moments due to loads are distributed.
    2. Compute the moments due to these forces.
    3. Combine the moments obtained in Steps 1 and 2 to eliminate the effect of the forces that prevented sidesway.

    Suppose the rigid frame in Fig. 5.77 is subjected to a 2000-lb horizontal load acting to the right at the level of beam BC. The first step is to compute the moment-
    influence factors (Table 5.6) by applying moments of 1000 at joints B and C,  assuming sidesway prevented.

    Since there are no intermediate loads on the beams and columns, the only fixed-end moments that need be considered
    are those in the columns resulting from lateral deflection of the frame caused by the horizontal load. This deflection, however is not known initially.
    So assume an arbitrary deflection, which¬†produces a fixed-end moment of¬†-1000M at the top of column CD. M is an unknown constant to be determined¬†fr ¬†the columns and hence are equal in AB to -1000M x 6‚ĀĄ2 = -3000M. The columnom the fact that the sum of the shears in the deflected columns must be equal to¬†the 2000-lb load. The same deflection also produces a moment of -1000M at the¬†bottom of CD [see Eq. (5.126)].

    From the geometry of the structure, furthermore, note that the deflection of B  relative to A is equal to the deflection of C relative to D. Then, according to Eq. (5.126) the fixed-end moments in the columns are proportional to the stiffnesses of  fixed-end moments are entered in the first line of Table 5.7, which is called a moment-collection table.

    from which M = 2.30. This value is substituted in the sidesway total in Table 5.7 to yield the sidesway moments for the 4000-lb load. The addition of these moments to the totals for no sidesway yields the final moments.
    This procedure enables one-story bents with straight beams to be analyzed with the necessity of solving only one equation with one unknown regardless of the number of bays. If the frame is several stories high, the procedure can be applied to each story. Since an arbitrary horizontal deflection is introduced at each floor or roof level, there are as many unknowns and equations as there are stories.
    The procedure is more difficult to apply to bents with curved or polygonal members between the columns. The effect of the change in the horizontal projection of the curved or polygonal portion of the bent must be included in the calculations.
    In many cases, it may be easier to analyze the bent as a curved beam (arch).
    (A. Kleinlogel, ‚Äė‚ÄėRigid Frame Formulas,‚Äô‚Äô Frederick Ungar Publishing Co., New¬†York.)

    Rapid Approximate Analysis of Multistory Frames

    Exact analysis of multistory rigid frames subjected to lateral forces, such as those from wind or earthquakes, involves lengthy calculations, and they are timeconsuming and expensive, even when performed with computers. Hence, approximate methods of analysis are an alternative, at least for preliminary designs and, for some structures, for final designs.
    It is noteworthy that for some buildings even the ‚Äė‚Äėexact‚Äô‚Äô methods, such as those¬†described in Arts. 5.11.8 and 5.11.9, are not exact. Usually, static horizontal loads¬†are assumed for design purposes, but actually the forces exerted by wind and earthquakes¬†are dynamic. In addition, these forces generally are uncertain in intensity,¬†direction, and duration. Earthquake forces, usually assumed as a percentage of the¬†mass of the building above each level, act at the base of the structure, not at each¬†floor level as is assumed in design, and accelerations at each level vary nearly¬†linearly with distance above the base. Also, at the beginning of a design, the sizes¬†of the members are not known. Consequently, the exact resistance to lateral deformation¬†cannot be calculated. Furthermore, floors, walls, and partitions help resist¬†the lateral forces in a very uncertain way. See Art. 5.12 for a method of calculating¬†the distribution of loads to rigid-frame bents.
    Portal Method. Since an exact analysis is impossible, most designers prefer a¬†wind-analysis method based on reasonable assumptions and requiring a minimum¬†of calculations. One such method is the so-called ‚Äė‚Äėportal method.‚Äô‚Äô
    It is based on the assumptions that points of inflection (zero bending moment) occur at the midpoints of all members and that exterior columns take half as much shear as do interior columns. These assumptions enable all moments and shears throughout the building frame to be computed by the laws of equilibrium.
    Consider, for example, the roof level (Fig. 5.78a) of a tall building. A wind load of 600 lb is assumed to act along the top line of girders. To apply the portal method, we cut the building along a section through the inflection points of the top-story columns, which are assumed to be at the column midpoints, 6 ft down from the top of the building. We need now consider only the portion of the structure above this section.
    Since the exterior columns take only half as much shear as do the interior columns, they each receive 100 lb, and the two interior columns, 200 lb. The moments at the tops of the columns equal these shears times the distance to the inflection point. The wall end of the end girder carries a moment equal to the moment in the column. (At the floor level below, as indicated in Fig. 5.78b, that end of the end girder carries a moment equal to the sum of the column moments.) Since the inflection point is at the midpoint of the girder, the moment at the inner end of the girder must the same as at the outer end. The moment in the adjoining girder can be found by subtracting this moment from the column moment, because the sum of the moments at the joint must be zero. (At the floor level below, as shown in Fig. 5.78b, the moment in the interior girder is found by subtracting the moment in the exterior girder from the sum of the column moments.)
    Girder shears then can be computed by dividing girder moments by the half span. When these shears have been found, column loads can be easily computed from the fact that the sum of the vertical loads must be zero, by taking a section around each joint through column and girder inflection points. As a check, it should be noted that the column loads produce a moment that must be equal to the moments of the wind loads above the section for which the column loads were computed.
    For the roof level (Fig. 5.78a), for example, -50 x 24 + 100  x 48 = 600 x 6.

    Cantilever Method. Another wind-analysis procedure that is sometimes employed is the cantilever method. Basic assumptions here are that inflection points are at the midpoints of all members and that direct stresses in the columns vary as the distances of the columns from the center of gravity of the bent. The assumptions are sufficient to enable shears and moments in the frame to be determined from the laws of equilibrium.
    For multistory buildings with height-to-width ratio of 4 or more, the Spurr modification
    is recommended (‚Äė‚ÄėWelded Tier Buildings,‚Äô‚Äô U.S. Steel Corp.). In this
    method, the moments of inertia of the girders at each level are made proportional to the girder shears.
    The results obtained from the cantilever method generally will be different from¬†those obtained by the portal method. In general, neither solution is correct, but the¬†answers provide a reasonable estimate of the resistance to be provided against¬†lateral deformation. (See also Transactions of the ASCE, Vol. 105, pp. 1713‚Äď1739,¬†1940.)

    Beams Stressed into the Plastic Range

    When an elastic material, such as structural steel, is loaded in tension with a gradually increasing load, stresses are proportional to strains up to the proportional limit (near the yield point). If the material, like steel, also is ductile, then it continues to carry load beyond the yield point, though strains increase rapidly with little increase in load (Fig. 5.79a).

    Similarly, a beam made of a ductile material continues to carry more load after the stresses in the outer surfaces reach the yield point. However, the stresses will no longer vary with distance from the neutral axis, so the flexure formula [Eq. (5.54)] no longer holds. However, if simplifying assumptions are made, approximating the stress-strain relationship beyond the elastic limit, the load-carrying capacity of the beam can be computed with satisfactory accuracy.

    Modulus of rupture is defined as the stress computed from the flexure formula for the maximum bending moment a beam sustains at failure. This is not a true stress but it is sometimes used to compare the strength of beams.
    For a ductile material, the idealized stress-strain relationship in Fig. 5.79b may be assumed. Stress is proportional to strain until the yield-point stress ƒy is reached, after which strain increases at
    a constant stress.
    For a beam of this material, the following assumptions will also be made:
    1. Plane sections remain plane, strains thus being proportional to distance from the neutral axis.
    2. Properties of the material in tension are the same as those in compression.
    3. Its fibers behave the same in flexure as in tension.
    4. Deformations remain small.

    Strain distribution across the cross section of a rectangular beam, based on these assumptions, is shown in Fig. 5.80a. At the yield point, the unit strain is y and the curvature y, as indicated in (1). In (2), the strain has increased several times, but the section still remains plane. Finally, at failure, (3), the strains are very large and nearly constant across upper and lower halves of the section.
    Corresponding stress distributions are shown in Fig. 5.80b. At the yield point, (1), stresses vary linearly and the maximum if ƒy . With increase in load, more and more fibers reach the yield point, and the stress distribution becomes nearly constant, as indicated in (2). Finally, at failure, (3), the stresses are constant across the top and bottom parts of the section and equal to the yield-point stress.
    The resisting moment at failure for a rectangular beam can be computed from the stress diagram for stage 3. If b is the width of the member and d its depth, then the ultimate moment for a rectangular beam is

    Since the resisting moment at stage 1 is My  ƒybd2 / 6, the beam carries 50% more moment before failure than when the yield-point stress is first reached at the outer surfaces.

    Tags: , , , , , , , | 9,886 views

Leave a Reply

Your email address will not be published. Required fields are marked *

+ 2 = ten

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>