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I am trying to understand some analytic expressions given in the handbook Roark's Formulas for Stress and Strain for bimetallic strip stresses. Namely, Roark's Formulas give the formulas

$$\text{Equivalent } EI = {w t_b^3 t_a E_b E_a \over 12(t_a E_a + t_b E_b)}K_1$$

Where $$K_1 = 4 + 6{t_a \over t_b} + 4\left({t_a \over t_b}\right)^2 + {E_a \over E_b}\left({t_a \over t_b}\right)^3 + {E_b \over E_a}{t_b \over t_a}$$

and stresses of the top and bottom layers of the bi-metal strip as:

$$\begin{alignat}{4} \sigma &= -&&{(\gamma_b - \gamma_a)\Delta T E_a \over K_1}\left[{3{t_a \over t_b} + 2\left({t_a \over t_b}\right)^2 - {E_b t_b \over E_a t_a}}\right] \\ \sigma &= &&{(\gamma_b - \gamma_a)\Delta T E_b \over K_1}\left[{3{t_a \over t_b} + 2 -{E_a \over E_b}\left({t_a \over t_b}\right)^3}\right] \end{alignat}$$

where subscripts $a$ and $b$ indicate the two different metal layers, $E$ are the Young's moduli, $t$ are the layer thicknesses, $w$ is the beam width, $\gamma$ is the coefficient of thermal expansion, and $\Delta T$ is the elevation in temperature.

While I understand the Timoshenko bi-metallic strip curvature derivation (S. Timoshenko, "Analysis of Bi-Metal Thermostats," J. Opt. Soc. Am. 11, 233-255 (1925)), which also appears on Wikipedia (https://en.wikipedia.org/wiki/Bimetallic_strip). I cannot understand how the formulas above can be derived from the Timoshenko derivation or what additional assumptions were made.

Roark's Formulas is completely unhelpful in not providing a clear reference in which this derivation is done. If someone could point out a reference I can use to obtain the result shown or can show how the derivation is done, I would be immensely grateful.

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  • $\begingroup$ The fomula you give for $K_1$ does't make sense. Take the trivial case where $t_a = t_b$ and $E_a = E_b$. That gives $K_1 = 21$ when it should obviously be $8$ (you just have a beam of thickkess $2t_a$.) Note: I haven't checked whether this is a typo by the OP or a mistake in Roark. $\endgroup$ – alephzero Jan 17 at 11:10
  • $\begingroup$ Yes there was a typo, thanks for catching that! K1 should now give a factor of 16 (which becomes 8 when divided by the 2 in the denominator from the $tE + tE$ term. $\endgroup$ – user19001 Jan 19 at 18:52

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