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Brockenbrough_Ch03.qxd  9/29/05  5:05 PM  Page 3.71



                                                           CONNECTIONS


                                                                                          CONNECTIONS  3.71

                                    Bearing/Tear-out.  Although we initially determined that six bolts are required, the following
                                  bearing/tear-out check may require an adjustment in this number.

                                    Bolt shear: φR v = 39.8 kips (reduced value calculated above)
                                    Bearing on splice plate: φR p = 0.75 × 2.4 × 1.0 × 0.5 × 58 = 52.2 kips
                                    Bearing on W14 × 99 flange: φR p = 0.75 × 2.4 × 1.0 × 0.780 × 65 = 91.3 kips
                                    Tear-out on splice plate: φR to = 0.75 × 1.2 × (2 − 0.5 × 1.0625) × 0.5 × 58 = 38.3 kips
                                    Tear-out on W14 × 99 flange: φR to = 0.75 × 1.2(2 − 0.5 × 1.0625)0.780 × 65 = 67.1 kips
                                  (Two more tear-out limit states are related to the spacing of the bolts, but these are obviously
                                  not critical.) Since the bolt shear value of 39.8 kips is greater than the splice plate tear-out, the shear
                                  bearing/tear-out limit state for the six bolts is
                                                 φR vpt = 4 × 39.8 + 2 × 38.3 = 236 kips > 208 kips  OK
                                    Whitmore Section. Calculate distance l w and design strength for limit state of yielding as follows:

                                                     l w = (6 tan 30)2 + 7.5 =14.43 in.
                                                   φR n = 0.9 × 36 × 14.43 × 0.5 = 234 kips > 208 kips
                                  Note that if l w > 14.5 in, 14.5 in would be used in the calculation of design strength.
                                    In addition to the checks for the bolts and splice plates, the column sections should also be
                                  checked for bearing and block shear rupture. These are not necessary in this case because t f = 0.780 >
                                  t p = 0.50, the edge distances for the column are the same as for the plates, and the column material
                                  is stronger than the plate material.
                                  Splices in Truss Chords.  Splices in truss chords must be designed for 50% of the chord load as an
                                  axial force, or 2% of the chord load as a transverse force, as discussed in Art. 3.5.3, even if the load
                                  is compression and the members are finished to bear. As discussed earlier, these splices may be posi-
                                  tioned in the center of a truss panel, and therefore must provide some degree of continuity to resist
                                  bending. For the tension chord, the splice must be designed to carry the full tensile load.
                                    Example of Design of Splice in Truss Chord.  Design the tension chord splice shown in Fig. 3.43
                                                                         7
                                  for a factored load of 800 kips. The bolts are A325X,  / 8 in in diameter, φr v = 27.1 kips. The load at
                                  this location is controlled by the smaller W14 × 90 shape, so the loads should be apportioned to
                                  flanges and the web based on this member. Thus, the flange load P f and the web load P w are

                                                              .
                                                         P =  0 710 ×14 520.  × 800  = 311 kips
                                                          f
                                                                  .
                                                                 26 5
                                                                      =
                                                                 2
                                                         p = 800 − × 311 178 kips
                                                          w
                                  The load path is such that the flange load P f passes from the W14 × 90 through the bolts into the
                                  flange plates and into the W14 × 120 flanges through a second set of bolts. The web load path is
                                  similar.
                                    FLANGE SPLICE DESIGN
                                    Member Limit States. Bolt pattern, although not a member limit state, must be established to
                                  check the chords. The minimum number of bolts required in double shear is   311  × 27.1 = 5.74. Try
                                                                                           2
                                  six bolts in two rows of three as shown in Fig. 3.43. This may need to be
                                  adjusted because of bearing/tear-out.
                                    Chord net section is checked to see if the holes in the W14 × 90 reduce its capacity below 800 kips.
                                  Assume that there will be two web holes in alignment with the flange holes.
                                                   A net = 26.5 − 4 × 1 × 0.710 − 2 × 1 × 0.440 = 22.8 in 2
                                                  φR net = 22.8 × 0.75 × 65 = 1111 kips > 800 kips  OK



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