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I'm wondering if the structural integrity of a solid cylindrical pylon is just as sturdy as a cubical one of the same volume given that they're both the same height.

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I assume that both pylons are composed of the same material. If you have a cylindrical pylon with height $h$ and volume $V$ then the radius can be calculated by $V = hA_{\text{base}} =h\pi r^2 \implies r = \sqrt{\frac{V}{h\pi}}$. For the cuboid pylon with a square cross-section with a length $a$ for the edge we can calculate $a$ by $V = h A_{\text{base}}=ha^2 \implies a=\sqrt{\frac{V}{h}}$.

Depending on what you mean by sturdy we can distinguish different kinds of loads.

Short summary: Circular structures are stiffer but more expensive to manufacture and sometimes not suited for the usage (e.g. as floor structure in buildings, where commonly I-beams are used). Read the part below for more detailed answer.

  1. Axial load (compressive and tensile in elastic regime) $F$: As $F=\sigma A_{\text{base}}$ both structures have same "sturdyness" because $A_{\text{base}}$ is the same for both cases, which implies $\sigma$ to be the same. But this is only true in theory, because we neglected the effects of the geometry differences and the finite length of the structures.
  2. Torsional load $T$: For this case we have $\tau_{\text{max}}=\frac{T}{W_T}$, in which $W_T$ is the torsional resistance torque. For a circular cross-section, we have $W_T=\frac{\pi}{2}r^3$ and for a square cross section we have $W_T=0.208 a^3$. If you calculate $$\frac{\tau_{\text{max,r}}}{\tau_{\text{max,a}}}=2\cdot 0.208 \sqrt{\pi}\approx 0.737$$ you will see that the maximal shear stress for the circular structure is preferable, as it is less. Similarly by using $\dfrac{d \varphi}{dx}=\frac{T}{GI_T}$, in which $G$ is the shear modulus (is same for both cases) and $I_T$ is the torsion constant. The integral of the derivative $\dfrac{d \varphi}{dx}$ gives the amount of rotation at length $x$. $I_{T,r}=\frac{\pi}{2}r^4$ and $I_{T,a}=0.141a^4$. Again consider the fraction $$\frac{(d \varphi/dx)_{r}}{(d \varphi/dx)_{a}}=\frac{I_{T,a}}{I_{T,r}}=2\cdot 0.141 \pi\approx 0.886.$$ This implies that the circular cross section will show less rotational deflection unter the same torsional load $T$.
  3. Consider the bending torque $T_y$. The bending line $w(x)$ can be determined for Bernoulli-Euler-beams by solving the following differential equation $$EI_yw''(x)=-T_y.$$ For circular cross-section $I_{y,r}=\frac{\pi}{4}r^4$ and for square cross-section $I_{y,a}=\frac{1}{12}a^4$. If you integrate the differential equation and apply boundary conditions you can calculate the maximal bending / deflection. You could also consider the stress due to $T_y$ which is given by $$\sigma = \frac{T_y}{I_y}z_{\text{height in beam cross section}}.$$ The fraction $\sigma_{r}$ by $\sigma_{a}$ is given as: $$\frac{\sigma_r}{\sigma_a}=\frac{I_{y,r}}{I_{y,a}}=\frac{3}{\pi}\approx 0.955.$$ Which indicates that the stress in the circular structure is more favorable.
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