## 525 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.

### 5.25.1 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.

Bolts. Capacity of one bolt in double shear (i.e., two shear planes) is rv = 2 x 21 x 0.4418 = 18.6 kips. Thus 60/18.6 = 3.23 or 4 bolts are required.

Angles. Gross area required = 60/21.6 = 2.78 in2. Try two angles 3 x 3 x V4 with area = 2.88 in2. Then net area An = 2.88 - 1 x 0.25 x 2 = 2.38 in2, and effective net area Ae = UAn = 0.85 x 2.38 = 2.02 in2. (The U factor accounts for shear lag, because only one leg of an angle is bolted.) The allowable load based on fracture of the net area is 2.02 x 0.5 x 58 = 58.6 kips < 60 kips. The angles are not adequate. Try two angles 3 x 3 x 5/16 with area = 3.55 in2. The net area is An = 3.55 - 1 x 0.3125 x 2 = 2.92 in2, Ae = 0.85 x 2.92 = 2.48 in2, and the allowable load based on net area fracture is 2.48 x 0.5 x 58 = 71.9 kips > 60 kips—OK.

Gusset. Assume that the gusset is 10 in wide at the top bolt. Thus the net width is 10 - 1 = 9 in, and since, for U = 0.85, this is greater than 0.85 x 10 = 8.5, use 8.5 in for the effective net width. For the gross area used to determine the yield limit state, the effective width of the plate at the top bolt can be determined using the Whitmore section described in the AISC ASD and LRFD manuals. For this example, the Whitmore section width is 2(9 x tan 30°) = 10.4 in. Because this is greater than the actual width of 10 in, use 10 in. Therefore, thickness required = 60/(10 x 21.6) = 0.28 in. For the effective net

t END

ELEVATION VIEW

FIGURE 5.22 Hanger supported by gussetplate connection with bolts.

t END

ELEVATION VIEW

FIGURE 5.22 Hanger supported by gussetplate connection with bolts.

area (fracture limit state), thickness required = 60/(8.5 X 29) = 0.24 in. Yield controls at 0.28 in. Use a 5/16-in gusset.

Additional checks required are for bearing and angle-leg tearout (block shear fracture).

Bearing. For a bolt spacing of 3 in and an edge distance of 1V2 in, the allowable bearing stress is 1.2Fu = 1.2 X 58 = 69.6 ksi, and the allowable bearing load is 69.6 X 0.3125 X 0.875 X 4 = 76.1 kips > 60 kips—OK.

Tearout. The cross-hatched region of the angle leg in Fig. 5.22 can tear out as a block. To investigate this, determine the tearout capacity of the angles: The shear area is

The tension area is

The tearout resistance is

Pto = 4.51 X 0.3 X 58 + 0.49 X 0.5 X 58 = 92.7 kips > 60 kips—OK

### 5.25.2 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 end-connection 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.

FIGURE 5.23 Pair of steel angles stitched together (a) with ring fill and high-strength bolts, (b) with stitch bar and welds.

FIGURE 5.23 Pair of steel angles stitched together (a) with ring fill and high-strength bolts, (b) with stitch bar and welds.

Welded connections have the disadvantage of requiring fitting. Where there are several identical pieces, however, jigs can be used to reduce fit-up time.

Example—AISC ASD. Suppose the hanger in the preceding example is to be connected to the gusset plate by a welded lap joint with fillet welds (Fig. 5.24). Since the two angles will now have no holes, no net-area fracture-limit-state checks need be made. As before, the gross area required is 60/21.6 = 2.78 in2. Try two angles 3 X 3 X V4, with gross area = 2.88 in2. Because only one leg of the angle is connected, shear lag is a consideration here, just as it was in the bolted case. Thus the effective area is 0.85 X 2.88 = 2.45 in2, and the capacity is 2.45 X 29.0 = 71.0 kips > 60 kips—OK. Note that the fracture allowable stress of 0.5Fu is used here. The limit state within the confines of the connection is fracture, not yield. Yield is the limit state in the angles outside the connection where the stress distribution in the angles becomes uniform.

The maximum-size fillet weld that can be used along the edge of the V4-in material is 3/i6 in. To provide space for landing the fillet welds along the back and edge of each angle, the minimum width of gusset plate should be 3 + 2(3/i6 + 5/i6) = 4 in, if 3/i6-in welds are used. A preliminary sketch of the joint (Fig. 5.24) indicates that with a minimum width of 4 in, the gusset plate will be about 6 in wide at the ends of the angles, where the load will reach 60 kips on transfer from the welds. For a 6-in width, the required thickness of plate is 60/(22 X 6) = 0.46 in. Use a 1/2-in plate.

For a z/2-in plate, the minimum-size fillet weld is 3/i6 in. Since the maximum size permitted also is 3/i6 in, use a 3/i6-in fillet weld. If an E70XX electrode is used to make the welds, the allowable shear is 2i ksi. The capacity of the welds then is 0.707 X 3/i6 X 2i = 2.78 kips per in. Length of weld required equals 60(2 X 2.78) = i0.8 in. Hence supply a total length of fillet welds of at least ii in, with at least 5V2 in along the toe and 51/2 in along the heel of each angle.

To check the V2-in-thick gusset plate, divide the capacity of the two opposite welds by the allowable shear stress: Gusset thickness required is 2 X 2.78/14.5 = 0.38 in < 0.50 in—OK. One rule of thumb for fillet welds on both faces opposite each other is to make the gusset thickness twice the weld size. However, this rule is too conservative in the present case because the gusset cannot fail through one section under each weld. A better way is to check for gusset tearout. The shear tearout area is Av = 6 X 0.5 X 2 = 6.0 in2, and the tension tearout area is At = 3 X 0.5 X 1 = 1.5 in2. Thus the tearout capacity is Pt0 = 6 X 0.3 X 58 + 1.5 X 0.5 X 58 = 148 kips > 60 kips—OK. This method recognizes a true limit state, whereas matching fillet-weld size to gusset-plate thickness does not.

### 5.25.3 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

J60k I 60k

FIGURE 5.24 Hanger supported by gusset-plate connection with welds.

J60k I 60k

FIGURE 5.24 Hanger supported by gusset-plate connection with welds.

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.

Note that in Fig. 5.25b though only the T is shown distorted either flange may distort enough to induce prying forces in the fasteners. Hence theoretical determination of Q is extremely complex. Research has shown, however, that the following approach from the AISC "Manual of Steel Construction—ASD'' gives reliable results: Let a = ratio of moment M2 at bolt line to moment 8M1, at stem line where 8 = ratio of net area (along the line of bolts) to gross area (at face of stem or angle leg), and a' = value of a for which the required thickness is a minimum or allowable tension per bolt is a maximum.

## Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

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