Page 528 - Shigley's Mechanical Engineering Design
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                 502   Mechanical Engineering Design
                  Figure 9–26             P
                                          2                                               P
                                          P
                  Double-lap joint.
                                          2
                                                                  (a)
                                                                 y
                                                 t o        l          l
                                                            2         2
                                                                                 t i
                                                  t          h      x
                                                  o
                                                        h


                                                                  (b)


                                          illustrate the principles.  The shear-stress distribution for the double lap joint of
                                          Fig. 9–26 is given by

                                                            Pω                     Pω     	  2E o t o − E i t i
                                                 τ(x) =            cosh(ωx) +
                                                       4b sinh(ωl/2)          4b cosh(ωl/2)  2E o t o + E i t i
                                                                (α i − α o ) 	Tω
                                                        +                           sinh(ωx)                (9–7)
                                                          (1/E o t o + 2/E i t i ) cosh(ωl/2)
                                          where


                                                                         G    1     2
                                                                   ω =           +
                                                                         h  E o t o  E i t i
                                          and E o , t o , α o , and E i , t i , α i , are the modulus, thickness, coefficient of thermal expansion
                                          for the outer and inner adherend, respectively; G, h, b, and l are the shear modulus,
                                          thickness, width, and length of the adhesive, respectively; and 	T is a change in tem-
                                          perature of the joint. If the adhesive is cured at an elevated temperature such that the
                                          stress-free temperature of the joint differs from the service temperature, the mismatch
                                          in thermal expansion of the outer and inner adherends induces a thermal shear across
                                          the adhesive.






                        EXAMPLE 9–7       The double-lap joint depicted in Fig. 9–26 consists of aluminum outer adherends and an
                                          inner steel adherend. The assembly is cured at 250°F and is stress-free at 200°F. The
                                          completed bond is subjected to an axial load of 2000 lbf at a service temperature of 70°F.
                                          The width b is 1 in, the length of the bond l is 1 in. Additional information is tabulated
                                          below:

                                                            G, psi      E, psi     , in/(in  . °F)  Thickness, in
                                                                                         −6
                                                                 6
                                           Adhesive         0.2(10 )                 55(10 )          0.020
                                                                                          −6
                                                                            6
                                           Outer adherend               10(10 )      13.3(10 )        0.150
                                                                                          −6
                                                                            6
                                           Inner adherend               30(10 )      6.0(10 )         0.100
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