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Example(3):
       Determine the maximum stress in the




                                                                                                   2200 lb
reinforced weld of the bracket plate in fig.
                                                                   2 in     2 in       4.5 in
(7.6) . Assume that the load varies from zero to
the maximum value
 Solution:
 The throat area of each weld is:
       A= 0.707×0.25× (6 – 0.5) ×1.2 = 1.16




                                                            6 in
    2
 in
 For each weld, r1=2 in, and the total polar                               O
 second moment of area is:
          l2                      5.5 2 2
J = 2A × 
         12  + r1 2  = 2 x1.16 × 
                     
                                            
                                    12 + 2  = 15.1   in                              ¼ in weld
                                         
The torque is
      T = 2200 × 6.5 = 14300 lb-in                                        Fig. (7.6)
The distance r in equation (7.4) is
      r = 2.75 2 + 2 2 = 3.35 in
Hence, the maximum torsional stress, by equation (7.4), is
            14300 × 3.35
       s=                = 3172.5       psi
               15.1
This stress is resolved into a vertical component
              3172.5 × 2
       sv =              = 1894   psi
                3.35
And a horizontal component
              3172.5 × 2.75
       sh =                 = 2604.3     psi
                  3.35
The direct stress has only a vertical component,
                 F   2200
       s vd =      =       = 948.3       psi
                2A 2 ×1.16
The total vertical stress is
      svt = 1894 +948.3 = 2842.3 psi
Therefore the resultant stress is
      s = 2842.3 2 + 2604.3 2 = 3855 psi
With a concentration factor of k’ = 2.7 from table (7.2) the significant stress is
      smax = 3855 × 2.7 = 10408.5 psi

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MET 304 Welded joints example-3-solution

  • 1. Example(3): Determine the maximum stress in the 2200 lb reinforced weld of the bracket plate in fig. 2 in 2 in 4.5 in (7.6) . Assume that the load varies from zero to the maximum value Solution: The throat area of each weld is: A= 0.707×0.25× (6 – 0.5) ×1.2 = 1.16 6 in 2 in For each weld, r1=2 in, and the total polar O second moment of area is:  l2   5.5 2 2 J = 2A ×  12 + r1 2  = 2 x1.16 ×     12 + 2  = 15.1 in ¼ in weld     The torque is T = 2200 × 6.5 = 14300 lb-in Fig. (7.6) The distance r in equation (7.4) is r = 2.75 2 + 2 2 = 3.35 in Hence, the maximum torsional stress, by equation (7.4), is 14300 × 3.35 s= = 3172.5 psi 15.1 This stress is resolved into a vertical component 3172.5 × 2 sv = = 1894 psi 3.35 And a horizontal component 3172.5 × 2.75 sh = = 2604.3 psi 3.35 The direct stress has only a vertical component, F 2200 s vd = = = 948.3 psi 2A 2 ×1.16 The total vertical stress is svt = 1894 +948.3 = 2842.3 psi Therefore the resultant stress is s = 2842.3 2 + 2604.3 2 = 3855 psi With a concentration factor of k’ = 2.7 from table (7.2) the significant stress is smax = 3855 × 2.7 = 10408.5 psi