Question 16.7: The Araldite for which data are given in Figure 16.24 is to ...
The Araldite for which data are given in Figure 16.24 is to be used for joining two sheets of metal. The joint should support loads of up to 180 N/mm width. If an overlap of 10 mm can be accommodated, determine the optimum sheet thickness.
Selecting a safety factor of 2 to accommodate uncertainty in the loading conditions, the design load becomes 360 N/mm.
The failure stress in the adhesive is given by
\tau=\frac{F / w}{L}=\frac{360 \times 10^{3}}{0.01}=36 MPaExamination of Figure 16.24 gives t/L = 0.16 for \tau=36 MPa.
If L = 0.01 m, then t=1.6 \times 10^{-3} m =1.6 mm
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