Question 9.7: The double-lap joint depicted in Fig. 9–26 consists of alumi...

In Eq. (9–7) the parameter ω is given by

τ (x) =\frac{Pω}{4b sinh(ωl/2)} cosh(ωx) +\left[ \frac {Pω}{4b cosh(ωl/2)} \left(\frac {2E_{o}t_{o} − E_{i} t_{i}}{2E_{o}t_{o} + E_{i}t_{i}}\right) +\frac {(α_{i} − α_{o})Tω}{(1/E_{o}t_{o} + 2/E_{i} t_{i} ) cosh(ωl/2)} \right] sinh(ωx)            (9–7)

w=\sqrt{ \frac {G}{h} \left(\frac {1}{E_{o}t_{o}} +\frac {2}{E_{i}t_{i}}  \right) }

=\sqrt{ \frac {0.2(10^{6})}{0.020} \left[\frac {1}{10(10^{6})0.15} +\frac {2}{30(10^{6})0.10}  \right] }= 3.65 in^{−1}

(a) For the thermal component, α_{i} − α_{o} = 6(10^{−6}) − 13.3(10^{−6}) = −7.3(10^{−6})  in/( in °F), T = 70 − 200 = −130°F,

τ_{th}(x) =\frac {(α_{i} − α_{o})Tω sinh(ωx)}{(1/E_{o}t_{o} + 2/E_{i} t_{i} ) cosh(ωl/2)}

τ_{th}(x) =\frac { −7.3(10^{−6})(−130)3.65 sinh(3.65x)}{ \left[ \frac {1}{10(10^{6})0.150} +\frac {2}{30(10^{6})0.100}\right] cosh \left[\frac {3.65(1)}{2}\right]}

= 816.4 sinh(3.65x)

The thermal stress is plotted in Fig. (9–27) and tabulated at x = −0.5, 0, and 0.5 in the table below.

(b) The bond is “balanced” (E_{o}t_{o} = E_{i} t_{i}/2), so the load-induced stress is given by

τ_{P}(x) =\frac {Pω  cosh(ωx)}{4b  sinh(ωl/2) }=\frac{2000(3.65) cosh(3.65x)}{4(1)3.0208} = 604.1 cosh(3.65x)                       (1)

The load-induced stress is plotted in Fig. (9–27) and tabulated at x = −0.5, 0, and 0.5 in the table below.

(c) Total stress table (in psi):