# Is it possible to achieve the same amount of stress for an equal amount of deformation but two different cross sections?

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.