Page 519 - Shigley's Mechanical Engineering Design
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                                                                         Welding, Bonding, and the Design of Permanent Joints  493



                             EXAMPLE 9–3       A specially rolled A36 structural steel section for the attachment has a cross section
                                               as shown in Fig. 9–19 and has yield and ultimate tensile strengths of 36 and 58 kpsi,
                                               respectively. It is statically loaded through the attachment centroid by a load of  F =
                                               24 kip. Unsymmetrical weld tracks can compensate for eccentricity such that there is
                                               no moment to be resisted by the welds. Specify the weld track lengths l 1 and l 2 for a
                                               5  -in fillet weld using an E70XX electrode. This is part of a design problem in which
                                               16
                                               the design variables include weld lengths and the fillet leg size.

                                    Solution   The y coordinate of the section centroid of the attachment is

                                                                           1(0.75)2 + 3(0.375)2
                                                                    y i A i
                                                                         =                   = 1.67 in
                                                              ¯ y =
                                                                     A i    0.75(2) + 0.375(2)
                                               Summing moments about point B to zero gives

                                                                M B = 0 =−F 1 b + F ¯y =−F 1 (4) + 24(1.67)
                                               from which
                                                                             F 1 = 10 kip
                                               It follows that
                                                                      F 2 = 24 − 10.0 = 14.0 kip
                                               The weld throat areas have to be in the ratio 14/10 = 1.4, that is, l 2 = 1.4l 1 . The weld
                                               length design variables are coupled by this relation, so l 1 is the weld length design vari-
                                               able. The other design variable is the fillet weld leg size h, which has been decided by
                                               the problem statement. From Table 9–4, the allowable shear stress on the throat τ all is
                                                                        τ all = 0.3(70) = 21 kpsi

                                               The shear stress τ on the 45° throat is
                                                                         F                 F
                                                               τ =               =
                                                                   (0.707)h(l 1 + l 2 )  (0.707)h(l 1 + 1.4l 1 )
                                                                        F
                                                                =               = τ all = 21 kpsi
                                                                   (0.707)h(2.4l 1 )
                                               from which the weld length l 1 is
                                                                              24
                                                                  l 1 =                  = 2.16 in
                                                                      21(0.707)0.3125(2.4)
                                               and
                                                                    l 2 = 1.4l 1 = 1.4(2.16) = 3.02 in



                       Figure 9–19                 F 1    l 1                   3 8  in

                                                     A
                                                                                2 in
                                               b
                                                                                           +      F = 24 kip
                                                   F 2                          2 in      y
                                                     B                          3
                                                            l 2                 4  in
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