外文翻译钢结构-青岛理工大学(共11页).doc
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1、精选优质文档-倾情为你奉上4-6 Inadequate Lateral Support As the distance between points of lateral support on the compression flange (l) becomes larger, there is a tendency for the compression flange to buckle laterally. There is no upper limit for l. To guard against the buckling tendency as l becomes larger, h
2、owever, the ASDS provides that be reduced. This, in effect, reserves some of the beam strength to resist the lateral buckling. Figure 4-10b shows a beam that has deflected vertically with a compression flange that has buckled laterally. The result is a twisting of the member. This is called lateral-
3、torsional buckling. For simplicity we refer to this buckling mode of the beam as lateral buckling. Two general resistances are available to counteract lateral buckling: torsional resistance of the cross section and lateral bending resistance of the compression flange. The total resistance to lateral
4、 buckling is the sum of the two. The ASDS conservatively considers only the larger of the two in the determination of a reduced . The ASDS, Section F1.3, establishes empirical expressions forfor the inadequate lateral support situation. Tension and compression allowable bending stresses are treated
5、separately. The tensionis always 0.60. Only the compression is reduced. For typical rolled shapes, this is of no consequence because the shapes are symmetrical and the lowerof the two values will control. Note that the provisions of this section pertain to members having an axis of symmetry in, and
6、loaded in, the plane of their web. They also apply to compression on extreme fibers of channels bent about their mayjor axis. The ASDS provides three empirical equations for the reduced compression. The mathematical expressions that give an exact prediction of the buckling strength of beams are too
7、complex for general use. Therefore, the ASDS equations only approximate this strength for purposes of determining a reasonable. The that is finally used is the larger of the values determined from the applicable equations. The first two, ASDS Equations (F1-6) and (F1-7), give the value when the late
8、ral bending resistance of the compression flange provides the lateral buckling resistance. The third, ASDS Equation (F1-8), gives when the torsional resiastance of the beam section provides the primary resistance to lateral buckling. In no case should be greater than 0.6for beams that have inadequat
9、e lateral support. The equations that will be applicable will depend on the value of the ratio l/, wherel=distance between points of lateral support for the compression flange (in.)=radius of gyration of a section comprising the compression flange plus one-third of the compression web area taken abo
10、ut an axis in the pane of the web (in.), as shown in Figure 4-11Here is a tabulated quantity for rolled shapes (see the ASDM, Part 1), and may be considered a slenderness ratio of the compression portion of the beam with respect to the y-y axis. The equations for are as follows:Where= a liberalizing
11、 modifying factor whose value is between 1.0 and 2.3 that accounts for a moment gradient over the span and a decrease in the lateral buckling tendency; may be conservatively taken as 1.0; see ASDS, Section F1.3, for details d= depth of cross section (in.)=area of compression flange ()Figure 4-12 dep
12、icts the decision-making process for the calculation of. Note that one will use ASDS Equations (F1-6) and (F1-8) or ASDS Equations (F1-7) and (F1-8). The larger resulting is used. Note that Table 5 of the Numerical Values section of the ASDS provides the following numerical equivalents for A36 steel
13、 (=36ksi):4-7 Design of Beams for Moment The basis for moment design is to provide a beam that has a moment capacity() equal to or greater than the anticipated maximum applied moment M. The flexure formula is used to determine a required section modulus S:The section modulus on which the selection w
14、ill be based is assumed to be the strong-axis section modulus. The Allowable Stress Design Selection Table ( Table) in the ASDM, Part 2, can be used to make this selection. It lists common beam shapes in order of decreasing section modulus. This table also lists the resisting momentof each section.
15、The value ofis calculated using an allowable bending stressof 23.8 ksi (or 23.76 ksi) rather than the rounded value of 24.0 ksi. This may cause some small inconsistencies in calculations and results.4-8:Beam Design for MomentBased on the foregoing examples,a general procedure may be established for
16、the design of beams for moment .1. Establish the condition of load,span,and lateral support.This is best done with a sketch.Establish the steel type.2. Determine the design moment.If necessary,complete shear and moment diagrams should be drawn.An estimated beam weight may be included in the applied
17、load.3. The beam curves should be used to select an appropriate section when possible.As an alternative, must be estimated and the required section modulus determined.4. After the section has been selected,recompute the design moment,including the effect of the weight of the section.Check to ensure
18、that the section selected is still adequate.5. Check any assumption that may have been made concerning or.6. Be sure that the solution to the design problem is plainly stated.4-9:Shear in BeamsExcept under very special loading conditions,all beams are subjected to shear as well as moment.In the norm
19、al process of design,beams are selected on the basic of the moment to be resisted and then checked for shear.Shear rarely controls a design unless loads are very heavy(and,possibly,close to the supports)and/or spans are very short.From strength of material,the shear stress that exists within a beam
20、may be determined from the general shear formula Where shear stress on a horizontal plane located with reference to the neutral axis(ksi) V=vertical shear force at that particular section(kips) Q=statical moment of area between the plane under consideration and the outside of the section,about the n
21、eutral axis()I=moment of inertia of the section about the neutral axis()b=thickness of the section at the plane being considered(in.)This formula furnishes us with the horizontal shear stress at a point,which,as shown in any strength of material text,is equal in intensity to the vertical shear stres
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