Why is the crosssection of the transmission shaft made into a circle?
Today, I will give you an analysis from the technical point of view.
1. Mechanical analysis of torsion
1. Torsion form
(1) Convention on torque symbols
Fig. 1 direction and symbol of torque
(2) Torsional deformation of circular section bar
After the round section shaft is twisted, the section remains flat, its shape and size remain unchanged, its radius remains the axis, and each section rotates relatively by only a small angle γ.
Fig. 2 torsional deformation of circular section bar
(3) Torsion of noncircular section bar
Fig. 3 torsional deformation of square bar
Free torsion:
When the crosssection of the bar is noncircular, the crosssection will warp during torsion deformation, and the warpage degree of adjacent crosssections is the same, so the length of all longitudinal fibers is not changed, and there is no normal stress but only shear stress on the crosssection.
Free torsion can be realized only when the two ends of the straight bar are subjected to external torque and the warpage of adjacent sections is not subject to external constraints.
Constrained torsion:
When the unequal straight bar is twisted, the torque changes along the length of the bar, or the end face is constrained and cannot be warped freely, the warpage degree of adjacent sections is different, and there is not only shear stress but also normal stress on the crosssection.
The normal stress caused by restrained torsion in a bar with solid section is usually small and can be ignored.
However, for thinwalled bars, this normal stress is often too large to be ignored.
2. Basic assumptions
(1) Plane hypothesis
After twisting, the circular section remains flat, its shape and size remain unchanged, its radius remains as the axis, and the sections rotate relatively by only a small angle γ.
This assumption holds only for the axis of circular section, but not for the axis of noncircular section.
The spacing between adjacent sections remains the same, except τzx= τzy, no normal stress;
σ x= σ y= σ z= τ xy=0.
The elasticity model is shown in Fig. 4.
Fig. 4 torsion elastic mechanical model of straight bar
(2) Membrane analogy
Prandt pointed out that the sag of thin membrane (liquid film) under uniform pressure is mathematically similar to the stress function in the torsion problem of straight bar with equal cross section.
Comparing the torsion bar with the membrane is helpful to solve the torsion problem.
There is a uniform film stretched on a horizontal boundary, as shown in Fig. 5.
The horizontal boundary has the same shape and size as the crosssectional boundary of a torsion bar.
When the film is subjected to a small uniform pressure, each point of the film will have a small sag.
If the plane where the boundary is located is the xy plane, the sag is z.
Due to the flexibility of the film, it can be assumed that it does not bear bending moment, torque, shear force and pressure, but only bears a uniform tensile force FT (similar to the surface tension of liquid film).
According to this analysis, the shear stress along any direction at a certain point on the crosssection of the torsion bar is equal to the slope of the film in the vertical direction at the corresponding point.
It can be seen that the maximum shear stress on the crosssection of the torsion bar is equal to the maximum slope of the membrane.
However, it should be noted that the direction of the maximum shear stress and the direction of the maximum slope are perpendicular to each other.
Using this assumption, the maximum shear stress and relative torsion angle of the noncircular section straight bar in Table 1 below can be obtained.
Fig. 5 Membrane analogy model
3. Calculation of torsional shear stress and torsional angle
(1) Solid circular shaft
Under assumptions 1 and 2, the mechanical properties of plastic materials in pure shear when the component materials are within the elastic range:
τ= G γ，γ Is the shear strain;
γ=φ R/L（ γ is the relative torsion angle of two sections at a distance of L;
φ is the corner of the end face of the torsion end, R is the outer radius of the circle, and L is the spacing between two sections).
Fig. 6 schematic diagram of torsion of bar with solid circular section
The shear stress at ρ on the circular section is:
Under the same torque condition, τρ is proportional to ρ on the crosssection of a circular crosssection bar, that is, the larger ρ is, the bigger τ is.
When ρ = R, the maximum shear stress at the edge of the circular section is obtained.
Since R and IP are only related to the geometric dimension of the section, they can be expressed as Wp = IP / R, which is called the torsional section modulus of the circular shaft, so τ max=T/WP.
It can be seen that the torsional shear stress of the bar is not directly related to its crosssectional area.
The torsional coefficient WP of the solid shaft with circular section is ≈ 0.2D3.
Torsion angle of round bar under torsion φ， GIP is the torsional stiffness of the circular section, reflecting the ability of the shaft to resist deformation.
The relative torsional angles of the two sections at a distance of L are:
Relative angle of twist:
Stiffness condition of circular shaft:
(2) Hollow circular shaft
The section torsional coefficient of hollow circular shaft is about: WP ≈ 0.2D^{3} (1 α ^{4}），0< α= d/D＜1.
When α= 0.8, the WP is 60% of the solid circular section, that is, under the same torque, the strength decreases by 40%, but under the same material and length, the weight difference is 2.8 times.
(3) Closed thinwalled tube
For a round pipe, when the wall thickness a of the round pipe is much smaller than the radius R0 (generally considered ≤ R0 / 10), it is called a thinwalled round pipe.
This kind of pipe can be of any shape and equal section.
Since it is a thinwalled pipe, it can be assumed that the shear stress is uniformly distributed in the whole wall thickness (t) to obtain an approximate solution.
According to the reciprocal shear stress, it can be concluded that the product of the average axial shear stress of all points on the pipe section and the pipe wall is equal, that is, the shear flow q is equal.
Since the q value is constant over the entire section, the shear stress is the largest at the minimum wall thickness.
When the pipe section is circular, Am = π R0 ², increasing the diameter of the cylinder can greatly reduce the shear stress.
4. Stress distribution of shaft cross section
Fig. 6 shear stress distribution of several common sections
2. Torsional failure mode
1. Destruction sequence
In the torsion test, the stress distribution of the specimen on the crosssection is uneven, the surface is the largest, and the more toward the center, the smaller.
Therefore, when twisting, the damage of the material is from the outside to the inside, and the crack first starts from the outermost layer of the round rod.
In engineering, torsion test is often used to check the surface defects of materials and the performance of surface hardening layer.
As shown in Fig. 7.
Fig. 7 torsion test of round bar sample
2. Plastic materials
For a circular shaft made of plastic materials such as lowcarbon steel, the surface of the shaft will yield first in the process of torsion, and then the circumference will be cut off along the section with the increase of torsional deformation.
Because the shear capacity of the material is lower than the tensile capacity and has the maximum shear stress on the crosssection, the fracture occurs on the crosssection and shows shear failure.
In engineering, the maximum shear stress on the outer edge of the crosssection is generally set to the shear yield limit of the material τs as the dangerous state, and based on this, the strength condition is established.
However, when the shear stress on the edge reaches the yield limit, the other parts are still in the linear elastic working state, and the round rod will not undergo obvious plastic deformation, that is, the torque can continue to increase.
When considering material plasticity, the ultimate torque (plastic torque) of a solid round rod is 1/3 larger than the yield torque (the result of simplified engineering calculation).
When the shear stress at the edge of the material crosssection reaches the shear yield limit of the material τs, with the increase of the torsional couple moment, the plastic region gradually expands inward, and the material at the edge of the crosssection begins to strengthen.
If the torsional couple moment continues to increase, the crack starts from the outermost layer of the round rod and is finally sheared along the crosssection.
As shown in Fig. 8.
Fig. 8 torsion test of round bar sample of plastic material
3. Brittle materials
For the round shaft made of brittle materials (such as cast iron) with tensile capacity lower than shear capacity, the deformation is very small during torsional failure, and it breaks on the helical surface at an angle of about 45 ° to the axis.
Because there is the maximum tensile stress on the inclined plane at 135 ° to the axis, when the maximum tensile stress on this section exceeds the tensile strength limit of the material, tensile failure occurs on this section.
As shown in Fig. 9.
Fig. 9 torsional test of brittle material round bar sample
4. Torsional failure of logs
The internal torque T received by the log rod not only generates a radial linear distribution of shear stress on the crosssection, but also generates a corresponding shear stress along the axial plane, resulting in cracking along the axial plane.
Because wood is an anisotropic material, the shear force parallel to the fiber along the axial direction is much smaller than the shear force in the section perpendicular to the fiber, thus showing the cracking form shown in Fig. 10.
Fig. 10 torsional failure of log
3. Torsional design of shaft
1. Torsion resistance of bars with different sections
According to the analysis of elasticity theory, the calculation formulas of maximum stress and torsion angle of the square section, triangular section and elliptical section are shown in the figure.
In all cases, the maximum shear stress exists at the position closest to the central axis on the section boundary line.
From the point of view of the closed thinwalled pipe, the position with the thinnest wall thickness relative to the central axis has the largest shear stress.
Fig. 11 calculation formula of torsional shear stress and relative torsional angle of different sections
Let the area of the circle, square, triangle and ellipse be S, and be affected by the same torque T.
Then, the side length of a square is a = S^{1 / 2}, and the side length of an equilateral triangle is a ≈ 2.3S ^{1 / 2}.
According to the maximum stress calculation formula given in the figure, under the same crosssectional area and torque, the maximum shear stress on the crosssection of an equilateral triangle is about 1.8 times that of a square;
For ellipse, when a = b, it is round, and a = 0.56S ^{1 / 2}, the maximum shear stress of square is about 1.32 times of that of circle;
Ellipse when a ≠ b, 1 > b / a= λ＞ 0, the ratio of the maximum shear stress of the ellipse to the maximum shear stress of the circle is λ S ^{– 2}, so the smaller the value of λ, the greater the shear stress.
Through the above comparison, it can be concluded that:
Under the condition of the same section and bearing the same torque, the maximum shear stress on the circular section is the smallest, and the torsion angle is also smaller than that of the noncircular section shaft.
Therefore, making the transmission shaft round has a natural comparative advantage in torsional mechanical performance.
Extending the above results to arbitrary crosssections, it can be proved that the circular crosssection shaft has the highest efficiency.
2. Estimate shaft diameter according to torque
When the length and span of the shaft are uncertain, the bearing reaction and bending moment cannot be obtained.
For multi fulcrum or less important shafts, the shaft diameter is often estimated according to the torque borne by the shaft, as shown in Table 1.
A value is the coefficient related to the material, which can be obtained by consulting the literature.
Table 1 torque check formula of shaft diameter
Axle type 
formula 
instruction 
solid shaft 
Where: d – calculate the diameter of the shaft at the section （mm） Trated torque transmitted by shaft （N·mm） T=9550000P/n Prated power transmitted by shaft （kW） nshaft speed (R / min) [T] – allowable shear stress of shaft （MPa） A – coefficient determined by [t], Vratio of inner diameter d0 to outer diameter D of the hollow circular shaft 

hollow axle 
3. Hollow shaft
Because the surface shear stress of circular crosssection shaft is large and the center is relatively small when it bears torsional load.
Therefore, removing some of the materials that do not play a full role in the center can effectively reduce the weight of the shaft and improve the bending resistance of the shaft.
However, whether shaft parts must be made hollow or not requires not only mechanical considerations, but also technological and manufacturing costs.
At the same time, the wall thickness should not be too thin, otherwise local folds will occur and the bearing capacity will be lost.
When the wall thickness δ of the cylinder is much smaller than the radius R0 (generally considered ≤ R0/10), it is called a thinwalled cylinder.
However, if the thinwalled tube has a longitudinal opening along the axis, its torsional resistance will be significantly reduced.
In order to improve its torsional stiffness and strength, diaphragm is usually added.
4. Stress concentration
A shaft is usually composed of different sections.
At the transition position of the section, the spatial mutation will cause stress concentration, which is also a common failure form of shaft parts.
How to select and determine the large diameter of two adjacent sections and the transition fillet can be referred to the literature.
5. Cylindrical coil spring
Cylindrical coil spring is a common part in mechanical engineering.
Its basic feature is that it has large elastic deformation and its axis is spiral.
In the design, for the spring with large load, its strength is usually considered;
But for the spring with little load, its deformation is generally considered;
Generally, the springs that are not important need only be selected according to the structural dimensions and the specifications.
For the design and calculation methods of springs, please refer to the literature, as well as GB / T1239 series standards, GB / T2089, DIN2089 and other standards.
4. Shear and tensile properties of materials
Under the action of static load, there is a certain relationship between the mechanical properties of materials in torsion and tensile, so [σ] of materials is used to determine the allowable shear stress[ τ]：
Material type  [σ]  [ τ]  [ τ] 
plastic material  1  0.5~0.7 [σ]  0.55 or 0.577 [σ] 
Brittle material  1  0.7~1.0 [σ]  0.8~1.0 [σ] 
It can be seen from the above table that the relationship between shear stress and normal stress given in the literature is different.
According to the ratio of shear stress to normal stress of several plastic materials provided in the literature, 0.5 ~ 0.7 [σ] is more appropriate.
The above relationship is only a rough estimation when there is no exact shear stress data.
To carry out accurate verification, it is necessary to obtain the specific torsional strength value of the material.