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I have a small tube with outer diameter (OD) $R_1$ mm and inner diameter (ID) $r_1$ mm. The tube is bent using a moment $M_1$ to create an angle $\theta_1$ with its plane of origin. The stress can then be calculated as $\sigma_1 = M_1/W_b$, where $W_b$ is the bend resistance $W_b = I/R$, where $I$ is the second moment of area and $r$ the distance from the centre axis to the edge of the tube.

Now consider that I have another tube with another OD and ID, $R_2$ and $r_2$ respectively. Is it possible to get the same amount of stress (or strain) for the same amount of bend/deformation? Considering that the two tubes are made out of the same material.

Mathematically speaking: $\sigma_1 = \sigma_2 \rightarrow \frac{M_1}{W_{b1}}=\frac{M_2}{W_{b2}}\tag{1}$

Since $W_b = I/R$ and $I=\frac{\pi(R^4-r^4)}{4}$, equation (1) can be written as:

$\frac{M_1R_1}{\frac{\pi(R_1^4-r_1^4)}{4}}= \frac{M_2R_2}{\frac{\pi(R_2^4-r_2^4)}{4}} \rightarrow \frac{4M_1R_1}{R_1^4-r_1^4} = \frac{4M_2R_2}{R_2^4-r_2^4} \rightarrow M_1R_1(R_2^4-r_2^4) = M_2R_2(R_1^4-r_1^4)\tag{2}$

Using Euler-Bernoulli beam theory, the angle $\theta$ can be expressed as:

$\theta = \frac{Ml}{EI}$ where $l$ is the lenght of the tube and $E$ is Young's modulus. Same angle results in:

$\theta_1 =\theta_2 \rightarrow \frac{M_1l}{EI_1} = \frac{M_2l}{EI_2} \tag{3}$

Since the lenght of the tube is kept the same and the material is the same, $E$ and $l$ are same and can be removed to yield:

$\frac{M_1}{I_1} = \frac{M_2}{I_2} \rightarrow \frac{M_1}{\frac{\pi(R_1^4-r_1^4)}{4}} = \frac{M_2}{\frac{\pi(R_2^4-r_2^4)}{4}} \rightarrow \frac{M_1}{R_1^4-r_1^4} = \frac{M_2}{R_2^4-r_2^4} \tag{4}$

Solving (4) for $M_2$ yields:

$M_2 = M_1\frac{R_2^4-r_2^4}{R_1^4-r_1^4}\tag{5}$

If the solution from equation (5) is inserted in (2) then the following result is reached:

$M_1R_1(R_2^4-r_2^4) = M_1\frac{R_2^4-r_2^4}{R_1^4-r_1^4}R_2(R_1^4-r_1^4) \implies (R_2^4-r_2^4)R_1 = (R_2^4-r_2^4)R_2 \tag{6}$

The result in (6) basically says that $R_1=R_2$, which implies that there is only one unique combination of cross section and stress which yields a specific deformation. Can this be true? This means that there is only one dimension of tube which will yield a specific stress for a certain bend.

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That is right. For the same material and the same deformation, the stress will only depend on distance distance from the center. So if you want to have same maximal stresses in that situation, the tubes have to have the same outside diameters. And when outside diameters are the same and the deformation is the same, the tubes have to have also same inside diameters.

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