142 X 750

By Eq. (5.14), the load per bolt due to shear is 48/12 = 4.00 kips. Consequently, the total load on the outermost bolt is the resultant:

The design strength of a 7/s in bolt is 0.75 X 0.6013 X 48 = 21.6 kips > 16.0 kips. The connection is satisfactory.

An ultimate-strength method that gives an accurate estimate of the strength of eccentrically loaded bolt groups may be used instead of the preceding procedure. The method assumes that fastener groups rotate about an instantaneous center. (This point coincides with the centroid of a group only when the moment arm l becomes very large.) Tables in the AISC ASD and LRFD manuals are based on this method.

Inelastic Method. Refer to Table 8-20 of the AISC LRFD Manual, Vol. II, Connections. With the moment arm ex = 15 in, bolt spacing s = 3 in, and number of bolts per row n = 6, interpolate between ex = 14 in and ex = 16 in to find the coefficient C = (3.99 + 3.55)/ 2 = 3.77. The connection strength then is 4>R = 3.77 X 21.6 = 81.4 kips > 48 kips, and is OK. The coefficient C indicates that, because of the eccentric loading, the 12 bolts of the connection are only as effective as 3.77 bolts in direct (concentric) shear. To compare methods, an equivalent C for the elastic method could be calculated as the ratio of the applied load to the resultant force on the outermost bolt, 48/16 = 3.00. Thus, for this example, the inelastic method gives a capacity 3.77/3.00 = 1.26 times that of the elastic method.

The plate should be checked for bending and shear. For the gross section, required bending strength is f = 48(15 - 2.75 - °.5) = fb 0.5(18)2/4 13.9 ksi

The strength of the gross section is

The plate is satisfactory for bending (yielding) of the gross section. For bending of the net section, from the AISC LRFD manual, the section modulus is Snet = 18 in3 and the required strength is

Fracture, rather than yielding, however, is the limit state for the net section. The fracture strength of a net section in tension is 0.75FuAe, where Fu is the specified tensile strength and Ae is the effective net area. Since bending induces a tensile stress over the top half of the bracket, and because the yield limit on the gross section has already been checked, it is reasonable to assume that the fracture (rupture) design strength of the bracket net section is 0.75 X 58 = 43.5 ksi > 32.7 ksi. The bracket is satisfactory for fracture of the net section. The shear on the gross section is f = n /8 10 = 5.33 < (0.9 X 0.6 X 36 = 19.4 ksi)—OK 0.5 X 18

The shear on the net section is fv = 0.5(18 - 68X 0.9375) = ^ < (a75 X a6 X 58 = 2° ksi)—OK

The plate is satisfactory for shear.

A plate bracket such as the one in Fig. 5.46 also can be connected to a support with fillet welds in shear. Design of the welds for such a connection can be performed by the classical elastic method, which is analogous to that for fasteners. For example, for the bracket in Fig. 5.47, the shear due to the 48-kip factored load induces a shear (kips per in) equal to the load divided by the total length of weld (weld A + weld B + weld C). The moment due to the load tends to rotate the welds about their center of gravity O. As a consequence:

The force at any point of a weld acts normal to the radius vector from O to the point. In Fig. 5.47, Pb, the force due to the moment of the 48-kip load about O, is normal to the radius vector OB.

The magnitude of Pb (kips per in) is proportional to the distance r from O.

FIGURE 5.47 Bracket with fillet welds in shear.

The resisting moment per inch of weld is proportional to the square of the distance from O.

The applied moment equals the total resisting moment, the sum of the resisting moments of all the welds in the group.

These consequences result in the relationship

where M = applied moment, in-kips

Pm = load due to M on the point of a weld most distant from the center of gravity of the weld group, kips per in J = sum of squares of distances of unit weld lengths from center of gravity of group, in3 (analogous to polar moment of inertia) c = distance of outermost point from center of gravity, in

Hence, when the moment applied to a weld group is known, the maximum stress in the welds can be computed from Eq. (5.16). This stress has to be added vectorially to the shear on the weld. The resultant stress must be less than the allowable capacity of the weld.

Depending on the weld pattern, the largest resultant stress does not necessarily occur at the outermost point of the weld group. Vectorial addition of shear and bending stresses may have to be performed for the most critical points in a group to determine the maximum.

Computation of weld stresses generally is simplified if the forces and distances are resolved into their horizontal and vertical components. Advantage can be taken of the fact that

where Iy = sum of squares of distances measured horizontally from center of gravity of weld group to unit lengths of welds, in3 Ix = sum of squares of distances measured vertically from center of gravity of weld group to unit lengths of welds, in3

Ix and Iy are analogous to moment of inertia.

Example—Welded Bracket Connection—AISC LRFD. Investigate the bracket connection in Fig. 5.47. The A36 steel bracket is to be connected with fillet welds made with E70XX electrodes to a building column. The bracket carries a 48-kip factored load 15 in from the center of the column web.

Elastic Method. Because of symmetry, the center of gravityO of the weld group is located vertically halfway between top and bottom of the 16-in-deep plate. The horizontal location of O relative to the vertical weld is obtained by dividing the moments of the weld lengths about the vertical weld by the total length of welds:

_ 2 x 7.5 x 7.5/2 56.2 , ni x = --7~— = —r— = 1.81 in

Welds A and B: 2 x 7.5(8)2 = 961 Welds A and B: ' = 70

By Eq. (5.16), the stress on the most distant point A in the weld group due to moment is

The vertical component of this load is

And the horizontal component is

The shear load on the welds is 48/(2 x 7.5 + 16) = 1.55 kips per in. Consequently, the total load on the outermost point is the resultant is

R = V(1.55 + 3.08)2 + (4.33)2 = 6.34 kips per in For a design stress of 0.75 x 0.60 x 70 = 31.5 ksi, the weld size required is

Instead of the elastic method, the following inelastic method based on the instantaneous center of rotation can be used. The tables for eccentrically loaded weld groups in the AISC manuals—ASD and LRFD—are based on this method.

Inelastic Method. From Fig. 5.47 and the AISC LRFD Manual, Vol. II, l = 16 in, kl = 7.5 in, and al = 11 + 7.5 - 1.81 = 16.69 in, from which a = 16.69/16 = 1.04 and k = 7.5/16 = 0.469. By interpolation in Table 8-42, coefficient C = 1.177. Hence the required weld size in number of sixteenths of an inch is

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