Contraction of Plate Assembly

 

Example:  Plate 1 is heated by an amount, ∆T, causing it to elongate to match the length of

plate 2, L.   The elongation of plate 1  is δM = 1 mm.   Both plates are then bonded together. 
When both plates return to original temperature find the strain, ε1, in plate 1.

                                                             

Data:  k1 = 39 x 103  N/mm,     k2  =  12 x 103  N/mm,    L  =  23 x 102  mm

 

 

Strategy:  (1)  Start with a free body diagram of the plates glued together.

                                          

                                  

 

(2)    Write equation of equilibrium. 

        Note: Statically indeterminate: 2 unknown forces with one equilibrium equation.

 

 

          →  ΣFx = 0     F1  - F2  = 0   and for the data:         F1  =   F2     -----------------  (1)

 

 

(3)  Examine compatibility of  displacements.  Final length of bonded plates is  L.

 

                                           

 

 

(4)  Apply force-extension relations:     F1  =  k1 (δM  -  δ) ,  F2  =  k2 ( δ)  -------- ( 2a, 2b)

 

Use (1) and (2)      F1  / F2   =  1  =   k1 (δM  -  δ) / k2 ( δ)     and solve for   δ

 

                  δ  =  k1 δM /  ( k1 + k2 )    then   δM  ˗  δ  =  contraction of  plate 1

 

and                                      δM  ˗  δ  =  k2 δM /  ( k1 + k2 )   

 

 

(5)  The strain in plate (1) is then   ε =  ( δM  ˗  δ ) /L  =   [ k2 / ( k1 + k2 )] [δM / L ]       (result)

 

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