Use workbook to check or manually generate Height/Span limit charts for coldformed c-sections or other sections, using AS1170.2 and Kleinlogel rigid frame formula (Frame III SCI steel designers manual 5th edition).
Structures formed of bars that are rigidly connected are referred to asframes, whilethose of bars that are pin connected aretrusses. Analytically, trusses are treated asbeing a special case of frames. For the frames of this chapter, it is assumed thatthere is no interaction between axial, torsional, and flexural deformations (i., theresponses are based on uncoupled extension, torsion, and bending theory). Formulas are provided for several simple frame configurations with simple load-ings. Also, structural matrices required for more complicated frames are listed. Manycommercially available general-purpose structural analysis computer programs canbe used to analyze complicated frames. Entries in most of the tables of this chapter give salient values of reactions, forces,and moments. Also, a moment diagram is shown. This moment can be used to cal-culate the bending stresses using the technical beam theory flexural stress formula.Formulas for buckling loads and natural frequencies are tabulated. Special attention is given to gridworks, which are flat networks of beams withtransverse loading. Collapse loads are provided for plastic design.
Kleinlogel Rigid Frame Formulas Free
A special case of frames is agridwork,orgrillage, which is a network of beamsrigidly connected at the intersections, loaded transversely. That is, a gridwork is anetwork of closely spaced beams with out-of-plane loading. It may be of any shapeand the network of beams may intersect at any angle. These beams need not beuniform. The gridworks treated here are plane structures (Fig. 13-5), with the beams lyingin one direction calledgirdersand those lying in the perpendicular direction calledstiffeners. Either set of gridwork beams can be selected to be the girders. In prac-tice, the wider spaced and heavier set is usually designated as girders, whereas thecloser spaced and lighter beams are stiffeners. For auniform gridwork, the girdersare identical in size, end conditions, and spacing. However, the set of stiffeners maydiffer from the set of girders, although the stiffeners are identical to each other. Thetreatment here is adapted from Ref [13]. For the formulas here, the cross section of the beams may be open or closed, al-though torsional rigidity is not taken into account. For closed cross sections this maylead to an error of up to 5%. Stresses in the girders and stiffeners can be calculatedusing the formulas for beams in Chapter 11.
For gridworks not covered by the formulas here, use can be made of a frameworkcomputer program. The structural matrices, including transfer, stiffness, and massmatrices, for a grillage are provided in Section 13. The sign convention of thetransfer matrix method for displacements and forces for the beams of Chapter 11apply to the gridwork beams here.
The direction of the reaction forces are shown in the figures of the configurations. Thesigns of the moments are shown in the moment diagrams. A bending moment is indicatedas positive when it causes tension on the inner side of the frame and compression on theouter side. Opposing moments are negative. The formulas in the table give the magnitudesof these quantities. The horizontal and vertical coordinate axes arexandy, respectively.vjkis the displacement of pointjin thekdirection.θjis the slope at j.
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.
The frame in Fig. 5.74 consists of four prismatic members rigidly connected togetherat 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).
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.
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 exact 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 portal method.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. 2ff7e9595c
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