# How to calculate stiffness to weight ratio of one big tube vs two small tubes (see drawing)?

How does one calculate the stiffness1 of (a) one big diameter tube and (b) two small diameter tubes that are a certain distance apart?

I have made a drawing that illustrates and specifies the question:

Assumptions and clarifications:

• The black lines can be thought of as infinitely stiff plates onto which the tubes (white) are welded to.
• I chose the tube diameters below so that the two small tubes weigh as much as the one big tube in order to make this calculation answer the stiffness to weight ratio question posed in the title.
• a = b = 20.45mm
• c = 60.5mm
• d = 40mm
• force lateral 1 = force lateral 2
• force torsional 1 = force torsional 2
• By "force torsional" I mean a force that is for example pressing into the plane on the right end if the short black line and pulling from the plane (towards the observer) on the left end of the short black line.
• The length (500mm) and the wall thicknesses (0.9mm) are equal for all tubes.

1 Lateral and torsional stiffness, could also be measured in displacement of the shorter black line in the drawing.

• Do the infinitely stiff plates remain horizontal or can they rotate with respect to each other? May 22, 2023 at 17:16
• I suggest they stay parallel to each other because that is close enough to reality and because it probably simplifies the answer. Thanks for asking. May 22, 2023 at 18:38

# Lateral stiffness

Assuming the plates stay parallel, you can use lateral deflection of beam end while keeping the 0 rotation there. Replacing one of the plates with a force-moment couple, $$F_1$$ and $$M_1$$, the bending moment along the beam of length $$L$$ will be following: $$M(x) = F_1\cdot (L-x)-M_1$$ Since how much the beam bends locally is proportional to the bending moment, i.e. (Bernoulli beam): $$\frac{d\varphi}{dx} = \frac{M(x)}{EI_y}$$ where:

• $$E$$ is Young's modulus
• $$I_y$$ is second moment of area with respect to $$y$$ axis, for thin tube $$I_y = \pi\cdot r^3\cdot t$$ you can get the function of rotation as a function of $$x$$: $$\varphi(x) = \varphi_0 + \int\limits_0^x \frac{M(x)}{EI_y} dx$$ The rotation at the start of the beam is zero $$\varphi_0 = 0$$ and after integration, you get the following expression: $$\varphi(x) = \frac{1}{EI_y} \cdot \left(F_1\cdot \left(L\cdot x-\frac{x^2}{2}\right)-M_1\cdot x\right)$$

Now it is important, that rotation at the end of the beam is also 0: $$\varphi(x=L) = \frac{1}{EI_y} \cdot \left(F_1\cdot \left(L\cdot L-\frac{L^2}{2}\right)-M_1\cdot L\right) = 0$$ From this, you get relationship between force and moment: $$M_1 = F_1\frac{L}{2}$$

Lateral deflection is proportional to local rotation: $$\frac{dw}{dx} = \varphi(x)$$ So the end deflection can be obtained by integration of rotation function with the relationship between end force and moment: $$w_L = w_0 + \int\limits_0^L \varphi(x) dx = 0+ \frac{1}{EI_y} \cdot \int\limits_0^L\left(F_1\cdot \left(L\cdot x-\frac{x^2}{2}\right)-F_1\frac{L}{2}\cdot x\right) dx$$ So at the end, the deflection due to lateral force will be: $$w_L = \frac{L^3}{12EI_y}\cdot F_1$$ And lateral stiffness of one tube is: $$k_F = \frac{12E}{L^3}I_y$$ For your 2 cases and loading by force $$F$$ this means:

• one tube: $$k_{F1} = \frac{12E}{L^3}\pi\cdot \frac{d^3}{8}\cdot t = \frac{3E}{2L^3}\pi\cdot d^3\cdot t$$
• two tubes: $$k_{F2} = 2\frac{12E}{L^3}\pi\cdot \frac{a^3}{8}\cdot t = \frac{3E}{L^3}\pi\cdot a^3\cdot t$$

# Torsional stiffness

Torsional stiffness $$k_{T1}$$ for single tube is trivial: $$k_{T1} = \frac{GI_p}{L}$$ Since $$G = \frac{E}{2(1+\nu)}$$ and $$I_p = 2I_y$$ (just for tube), the stiffness will be: $$k_{T1} = \frac{E\cdot \pi\cdot d^3\cdot t}{16(1+\nu)L}$$

In case of two tubes, you need to take into account their torsional stiffness, but also lateral stiffness, since twisting the couple by angle $$\vartheta$$ will twist individual section by the same angle, but also laterally displace ends of the beams by $$\vartheta\cdot (c+a)/2$$, which would be caused by force from moment $$M = F\cdot (c+a)$$. Using only this lateral deflection, we can write: $$\frac{k_{F2}}{2} \cdot \frac{c+a}{2}\cdot \vartheta = \frac{M}{c+a}$$ $$\frac{(c+a)^2}{4} k_{F2} \cdot \vartheta = M$$

So the total torsional stiffness will be:

$$k_{T2} = \frac{E\cdot \pi\cdot a^3\cdot t}{8(1+\nu)L} + \frac{(c+a)^2}{4} k_{F2}$$

# Stiffness to weight ratio

I am not sure if stiffness to weight ratio is meaningful for this case, but since the both cases have the same weight, relative stiffness difference will be the same. Using relation between $$a$$ and $$d$$; $$d = 2a$$ (from thin shell assumption), you can calculate how much is the second case stiffer compared to the first one:

• lateral sitffness: $$\frac{k_{F2}}{k_{F1}} = \frac{1}{4}$$
• torsional stiffness: $$\frac{k_{T2}}{k_{T1}} = \frac{1}{4}+\frac{3(c+a)^2}{2}\cdot\frac{1+\nu}{L^2}$$

In conclusion, lateral stiffness of single tube is $$4\times$$ greater than for two tubes. Torsional stiffness is more complicated, but if we focus only on tube twisting, the case of one tube would also be $$4\times$$ greater. However, the two tube case has additional stiffness resulting from lateral stiffness, which you can calculate for your particular set of parameters.

• Thanks! "L" stands for angular momentum? May 22, 2023 at 18:57
• No, $L$ is the length, I will add it to the answer. May 22, 2023 at 20:15
• I think I understood everything you wrote now. I will answer my own question here with the specific numbers I provided in the question and draw a few general conclusions, all based on your answer. Thank you! May 23, 2023 at 6:31
• Is the unit for the stiffness equations in your answer N/mm (assuming I use mm in the formulas for a, c and d)? May 23, 2023 at 13:20
• I saw your section on "stiffness to weight ratio" after I also posted an answer (informed by yours). In my calculation the one tube is laterally 3.7417 times (0.15877593 / 0.042434) stiffer than the two tubes of the same weight, not 4 times as you have derived. Do you see a mistake I may have made? May 23, 2023 at 13:36

Caveat: I asked the question and give an answer here myself that takes the specific numbers given in my question and applies them to the formulas in the answer given by Tomas Letal. I do not have the capability to verify whether the theory behind Tomas' answer is correct or not. I couldn't figure out how to display the equations below with MathJax like Tomas did :/

Lateral stiffness

Lateral stiffness of the one tube: ((3 * 100)/(2 * 555^3)) * pi * 40^3 * 0.9 = 0.15877593021519

Lateral stiffness of the two tubes: ((3 * 100)/(555^3)) * pi * 20.45^3 * 0.9 = 0.04243406375145

To achieve the same lateral stiffness as the one 40mm diameter tube the two tubes would each have to have a diameter of about 31.75mm: ((3 * 100)/(555^3)) * pi * 31.748^3 * 0.9 = 0.15877561455371. Two 31.75mm diameter tubes weigh 1.578 times as much as the one 40mm diameter tube (assumption: same wall hickness) and the "weight to torsional stiffness ratio" is 1.578 accordingly.

Torsional stiffness

Torsional stiffness of the one tube: (100 * pi * 40^3 * 0.9) / (16 * (1 + 1) * 555) = 1018.89491467777

Torsional stiffness of the two tubes: ((100 * pi * 20.45^3 * 0.9) / (8 * (1 + 1) * 555))+(((60.5+20.45)^2)/4)*((3 * 100)/(555^3)) * pi * 20.45^3 * 0.9 = 341.823914090528

To achieve the same torsional stiffness as the one 40mm diameter tube the two tubes would each have to have a diameter of about 29.0mm: ((100 * pi * 29^3 * 0.9) / (8 * (1 + 1) * 555))+(((60.5+29)^2)/4)*((3 * 100)/(555^3)) * pi * 29^3 * 0.9 = 1018.89. Two 29.0mm diameter tubes weigh 1.43745 times as much as the one 40mm diameter tube (assumption: same wall thickness) and the "weight to torsional stiffness ratio" is 1.43745 accordingly.

Note that "weight to stiffness ratio" is different from "stiffness to weight ratio". In my question I asked for the "stiffness to weight ratio". Now I have calculated the "weight to stiffness" ratio. Not a perfect answer to my question but also helpful.