Ok

The tensile capacity is

Pt = 0.5 FhAe = 0.5 x 65 x 6.04 = 196.5 kips ^ governs

The tensile capacity based on yielding condition is

Therefore, tension capacity of 196.5 kips, as calculated for fracture condition along c-c, controls the design.

7.1.1.1.2. Plate in Tension, Welded Connection.

Given. A 3/4-in.-thick x 4-in.-wide plate welded to a gusset plate with two longitudinal welds (Fig 7.1c). The plate material is Fy = 36 ksi and Fu = 58 ksi. Assume that block shear does not control the design and that the welds are adequate.

Required. Calculate the tensile force that can be applied to the plate.

Solution. The design parameters are width of plate W = 4 in., thickness t = 0.75 in., and weld length l = 5 in.

The reduction coefficient U that accounts for the effects of eccentricity and shear lag is U = 0.87 (ASCE-B3)

The gross area of the plate is

Ag = 0.75 x 4 = 3 in.2 The effective net area is Ae = UAg

The corresponding tensile capacity is

The tension capacity for yielding condition is

Pt = 0.6 FyAg = 0.6 x 36 x 3 = 64.80 kips ^ governs

For other types of tension connections using rolled structural sections, such as W, M, and S, and angle shapes, the designer is referred to AISC section B3.

7.1.2. Members Subject to Bending

7.1.2.1. Lateral Stability

Consider a uniformly loaded continuous wide-flange beam as shown in Fig. 7.2. The beam segment between the points of contraflexure is subjected to positive bending with the top portion in compression throughout this region, acting in a manner similar to a column. Unless there are closely spaced restraints, i.e., ly < Le, the compression portion of the beam has a tendency to buckle laterally at some value of critical moment, Mcr, analogous to the critical load, Pcr, at which the column would buckle. The mode of lateral buckling is in torsion, partly due St. Venant's twisting, and partly due to warping torsion, the latter induced by the bending of beam flanges in opposite directions. Deep I-shaped open sections typically have large values for warping moment of inertia Iw. Consequently, the buckling mode for such beams is dominated by flange bending. On the other hand, for shallow I-beams, the St. Venant's torsion dominates the torsional response. These two considerations, the warping torsion and the St. Venant's torsion, lead to the two AISC equations, discussed presently for determining the allowable stress Fb for both compact and noncompact I-shaped sections with unbraced length ly > Lc.

7.1.2.2. Compact, Noncompact, and Slender Element Sections

There are two main categories of beams, compact and noncompact. Compact beams by virtue for their special controls on their geometry are particularly stable. Compact section criteria are based on the yield strength of steel, the type of cross section, the ratios of

Figure 7.2. Lateral torsional buckling of beams: (a) continuous beam with uniformly distributed load; (b) bending moment diagram showing positive bending region; (c) points of contraflexure.

width to thickness of the elements of cross section. Members, whether they are rolled shapes or shapes made from plates, are compact if they fulfill the criteria. To meet all criteria, the member must meet certain stringent limits of bfltf and h/tw ratios, have unsupported length of compression flange less than Lc, and be bent about its major axis. Nonfulfillment of any of the compact criteria will degrade the member to noncompact. If the width: thickness ratios exceed the limiting width: thickness ratios of noncompact elements, then it is classified as a slender element section.

7.1.2.3. Allowable Bending Stresses

To obtain the allowable bending stress, the following criteria are used. For all I- and C-sections, the allowable major direction bending stress is computed based on the compactness criteria and the laterally unbraced compression flange length, ly. If ly is less than

76b, and

20,000

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