Page 3 - Mechanics of Structures – Chapter 1
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Mechanics of Civil Engineering Structures – Chapter 5



               7.2.1 Single Shear

               "Single shear" refers to the fact that there is a single plane of shear force through the fastener -
               an example would be a tie rod end stud. Even with the taper to locate the parts, there is a single
               shear  plane  -  it  exists  between  the  tie  rod  end  housing  and  the  knuckle  surface  (typically
               considered to be even with the lower plane of the knuckle surface.)

                                                 ,  
                                      ,     =            n = no. of planes = 1
                                                     


               7.2.2 Double Shear

               "Double shear" would mean that there are two shear planes through the fastener, or that the
               total shear force is "split" into two locations on the fastener body. It's most common to see hitch
               pins loaded in double shear - a good example of this would be the cross pin for your trailer hitch.

                                                               ,  
                                                     ,     =        , n =  ≥ 2
                                                                   

                                                               ,  
                                                     ,     =       , [Rectangular cross section]
                                                       2        

                                                                ,  
                                                     ,     =       ,  [circular cross section]
                                                      2        2/4


               Example 1


                       Two pieces of steel plates connected with rivets with a diameter of 20 mm. If the tensile
                       force applied was 25 kN, estimate the shear stresses in the rivets.



                                                                                25kN
                                                                                    30k
                                   25 kN


                                                =  F / A

                                                         ( ) 2
                                                          20
                                               =
                                                 25kN  /
                                                           4
                                               =0.08 kN/mm  2






               NHO JKA PSAS                                                                             Page 3
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