I want to replace steel plates that are being used as trim tabs on a boat with 90-10 cupronickel. The main design parameter is to make sure the plate has no less flex than the part being replaced. In other words if we were to take strips of both materials, fix them at one end and put a weight on the other, the replacement material should not bend down any more than the original material.

The steel I assume probably has a tensile strength of around 50,000 PSI and the cupronickel I guess is maybe half that. So, in that case do I just make the cupronickel sheet double the thickness of the steel, or is it more complicated than that? What is the relevant formula or calculation?

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    $\begingroup$ for deflection you need to scale vs modulus of elasticity, not strength. look up simple beam bending (cantilevered with end load) $\endgroup$ – Pete W Apr 26 at 17:11

For a rectangular beam under a load, P, at the end the deflection is

$$ \delta= \frac{PL^3}{3EI}$$

therefore if you have the E as half you need to double the I of the part.

In rectangular beams and roughly rectangular tabs:

$$ I= \frac{BH^3}{12}$$

So you ned a tab thicker by the ratio of $T_{new}= T_{old}*\sqrt[3]{2}$


if we are concerned with a yield stress half of steel we have to try the section . modulus:

$$ S= \frac{BH^2}{6}$$

then the increase in thickness will be

$$Tn= T*\sqrt{2}$$

but now as a double check we need to check the deflection of new tab too.

  • $\begingroup$ OP is guessing tensile strength is maybe half, not Young’s modulus. Perhaps you should define the terms as E isn’t clear to a nonengineer. $\endgroup$ – Eric S Apr 26 at 21:35

$I.$ - Assume a cantilever beam and deflection control

$\Delta = P*L^3/3E*I, I = b*t^3/12$, where $\Delta$ is the deflection of a cantilever beam loaded on its free end, P is a unit force, L is the beam length, E is the elastic modulus of the plate, I is the moment of inertia, b is the beam width, and t is the beam depth (plate thickness). Let's plug I into the equation of deflection,

$\Delta = 4P*L^3/b*t^3*E$. In order for the flexibility to remain the same, both beams must be deflected to the same amount, that is

$\Delta_o = \Delta_1$ --->$4P*L^3/b*t_o^3*E_o = 4P*L^3/b*t_1^3*E_1$, the subscription "o" denotes the original plate, and "1" for the new plate. After eliminate the identical parameters of both beams, we get the equation

$t_o^3*E_o = t_1^3*E_1 $, solving for $t_1$ we get

$t_1 = [t_o^3*(E_o/E_1]^{1/3}$

If $E_1 =E_o/2$, $E_o/E_1 = 2$, then

$t_1 = [t_o^3*(2)]^{1/3} = (2)^{1/3}*t_o \approx 1.26*t_o$

$II.$ - Check strength requirement

$M = f_y*S$, in which $f_y$ is the yield strength, and

$S$ is the section modulus, $S = b*t^2/6$,

plug $S$ into the moment equation,

we get $M = f_y*b*t^2/6$.

Let $M_o = M_1$, after eliminating the common constants, we get

$f_{yo}*t_o^2 = f_{y1}*t_1^2$. Solving for $t_1$

$t_1 = t_o^2*\sqrt{(f_{yo}/f_{y1})}$

Check required plate thickness if $f_{y1} = f_{yo}/2$, so

$f_{yo}/f_{y1} = 2$, then

$t_1 = t_o*\sqrt{2} \approx 1.414*t_o$

$III.$ Conclusion

By comparison, $t_1$ derived from the deflection consideration ($1.26*t_o$) is less than that derived from strength concern ($1.414*t_o$), so the latter case governs (if I didn't make mistake in the calculations)./


Another way to do it is Monel , 70 Ni : 30 Cu. It should be as available as cupronickel with tensile of 80,000 psi ( depending on thickness ). Essentially the same or stronger than a hull steel so the same thickness as steel. Also available is 70 Cu : 30 Ni cupronickel , tensile strength about 65,000 psi ( depending on thickness ). Again a direct substitute for steel at the same thickness. Generally these materials are more available along a coast . I suggest calling various vendors to determine availability of the quantity and sizes you need. These two common alloys could be directly substituted for steel. A benefit you may not need is both alloys have excellent low temperature toughness , unlike hull steels. As I remember getting tough hull steel like ASTM A131 Grade E could be limited ( likely an obsolete grade today).


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