Hanger Connections

In buildings, end connections for hangers should be designed for the full loads on the hangers.
In trusses, however, the AISC specification requires that end connections should develop not only the design load but also at least 50% of the effective strength of the members. This does not apply if a smaller amount is justified by an engineering analysis that considers other factors, including loads from handling, shipping, and erection. This requirement is intended only for shop-assembled trusses where such loads may be significantly different from the loads for which the trusses were designed.
In highway bridges, connections should be designed for the average of the calculated stress and the strength of the members. But the connections should be capable of developing at least 75% of the strength of the members.
In railroad bridges, end connections of main tension members should have a strength at least equal to that of the members. Connections for secondary and bracing members should develop at least the average of the calculated stress and the strength of the members. But bracing members used as ties or struts to reduce the unsupported length of a member need not be connected for more than the flexural strength of that member.
When a connection is made with fasteners, the end fasteners carry a greater load than those at the center of the connection. Because of this, AISC and AASHTO reduce fastener strength when the length of a connection exceeds 50 in.

Bolted Lap Joints

Tension members serving as hangers may be connected to supports in any of several ways.
One of the most common is use of a lap joint, with fasteners or welds.
Example AISC ASD. A pair of A36 steel angles in a building carry a 60-kip vertically suspended load (Fig. 5.22). Size the angles and gusset plate, and determine the number of 7⁄8-in-diameter A325N (threads included) bolts required.

Welded Lap Joints

For welded lap joints, standard specifications require that the amount of lap be five times the thickness of the thinner part joined, but not less than 1 in. Lap joints with plates or bars subjected to axial stress should be fillet-welded along the end of both lapped parts, unless the deflection of the lapped parts is sufficiently restrained to prevent the joint from opening under maximum loading.
Welded end connections have the advantage of avoiding deductions of hole areas in determining net section of tension members.
If the tension member consists of a pair of angles required to be stitched together, ring fills and fully tensioned bolts can be used (Fig. 5.23a). With welded connections, welded stitch bars (Fig. 5.23b) should be used. Care should be taken to avoid undercutting at the toes of the angles at end-connection welds and stitch welds. In building design endconnection welds may be placed equally on the toe and heel of the angles, ignoring the small eccentricity. The welds should be returned around the end of each angle.

Fasteners in Tension

As an alternative to lap joints, with fasteners or welds in shear, hangers also may be supported by fasteners in tension. Permissible tension in such fasteners equals the product of the reduced cross-sectional area at threads and allowable tensile stress. The stress is based on bolts with hexagonal or square heads and nuts. Flattened-or countersunk-head fasteners, therefore, should not be used in joints where they will be stressed in tension.
Bolts designed for tension loads usually have a deliberately applied pretension. The tension is maintained by compression in the connected parts. A tensile force applied to a fastener relieves the compression in the connected parts without increasing the tension in the fastener.
Unless the tensile force is large enough to permit the connected parts to separate, the tension in the fastener will not exceed the pretension.
Generally, the total force on a fastener in tension equals the average force on all the fasteners in the joint plus force due to eccentricity, if present. Sometimes, however, the configuration of the joint produces a prying effect on the fasteners that may be serious and should be investigated.
Figure 5.25a shows a connection between the flange of a supporting member and the flange of a T shape (tee, half wide-flange beam, or pair of angles with plate between), with bolts in tension. The load P is concentrically applied. If the prying force is ignored, the average force on any fastener is P/ n, where n is the total number of fasteners in the joint.
But, as indicated in Fig. 5.25b, distortion of the T flange induces an additional prying force Q in the fastener. This force is negligible when the connected flanges are thick relative to the fastener gages or when the flanges are thin enough to be flexible.

Welded Butt Joints

A hanger also may be connected with a simple welded butt joint (Fig. 5.27). The allowable stress for the complete-penetration groove weld is the same as for the base metal used.
Examples AISC ASD. A bar hanger carries a 60-kip load and is supported through a
complete-penetration groove weld at the edge of a tee stub.
For A36 steel with allowable tensile stress of 22 ksi, the bar should have an area of 60⁄22 = 2.73 in^2. A bar 51⁄2 x 1⁄2 with an area of 2.75 in^2 could be used. The weld strength is equal to that of the base metal. Hence no allowance need be made for its presence.
Example AASHTO ASD. Suppose, however, that the hanger is to be used in a highway bridge and the load will range from 30 kips in compression (minus) to 60 kips in tension (plus). The design then will be governed by the allowable stresses in fatigue. These depend on the stress range and the number of cycles of load the structure will be subjected to.
Design is to be based on 100,000 cycles. Assuming a redundant load path and the detail of Fig. 5.27, load category C applies with an allowable stress range SR = 35.5 ksi, according to AASHTO. For A36 steel, the maximum tensile stress is 20 ksi. At 60-kip tension, the area required is 60⁄20 = 3 in^2. A bar 5 x 5⁄8 could be tried. It has an area of 3.125 in^2. Then the maximum tensile stress is 60/3.125 = 19.2 ksi, and the maximum compressive stress is 30/3.125  9.6 ksi. Thus the stress range is 19.2 - (-9.6) = 28.8 ksi The 5  5⁄8 bar is satisfactory.
As another example, consider the preceding case subjected to 2 million cycles. The allowable stress range is now 13 ksi. The required area is [60 - (-30)] /13 = 6.92 in^2. Try a bar 5  11⁄2 with area 7.5 in2. The maximum tensile stress is 60/ 7.5  8.0 ksi (20 ksi), and the maximum compressive stress is 30/ 7.5  4.0 ksi. Thus the actual stress range is 8.0  (4.0) = 12.0 ksi < 13.0 ksi OK. The 5 x 11⁄2 bar is satisfactory.

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