1. Torsional mechanical analysis
1. Twist shape
(1) Convention on torque symbols
Fig. 1 direction and torque symbol
(2) Torsional deformation of circular section bar
After twisting a shaft with a circular cross section, the shape and size of the section remain the same and it remains flat. The section radius remains the axis around which the section is twisted, and each section only rotates at a small angle γ to each other.
Fig. 2 Torsional deformation of the circular section bar
(3) Non-circular section bar twist
Fig. 3 Torsional deformation of square bar
Free Twist:
When a bar has a non-circular cross section, it will warp during torsional deformation. The degree of warping of adjacent cross sections will be the same, which means that the length of all longitudinal fibers of the bar will not change. In this scenario, there will be no normal stress in the cross section, only shear stress.
To obtain free torsion, both ends of the straight bar must be subjected to external torque and the bending of adjacent sections must not be externally restricted.
Restricted Twist:
When a non-uniform straight bar is twisted, the amount of torque applied changes along the length of the bar. If one end of the bar is fixed and cannot move, the degree of warping of adjacent sections of the bar will be different. In addition to shear stress, there will also be normal stress in the cross section of the bar.
Normally, the normal stress caused by restrained torsion in a solid bar is small and can be neglected. However, for thin-walled bars, this normal stress is often too great to ignore.
2. Basic assumptions
(1) Plan hypothesis
After twisting, the circular section remains flat and its shape, size and radius remain unchanged. The sections rotate relative to each other only at a small angle γ. However, this assumption only applies to the axis of the circular section and not to the axis of non-circular sections.
The spacing between adjacent sections remains the same, except when τzx = τzy, which indicates that there is no normal stress.
σ x= σ y= σ z= τ xy=0.
The elasticity model is shown in Fig.
Fig. 4 Elastic mechanical model of straight bar torsion
(2) M membrane analogy
Prandtl pointed out that the curvature of a thin liquid film, also known as a membrane, under uniform pressure is mathematically similar to the stress function in the torsion problem of a straight bar with equal cross-section.
Comparing the torsion bar to the membrane can be helpful in solving the torsion problem.
In Figure 5, there is a uniform film stretched over a horizontal boundary, which has the same shape and size as the cross-sectional boundary of a torsion bar.
When a small uniform pressure is applied to the film, each point on the film will undergo a small curvature.
If the plane where the limit is located is the xy plane, the arrow can be represented by z.
Due to the flexible nature of the film, it is assumed that it cannot withstand bending moment, torque, shear force or pressure. It only supports a uniform tensile force FT, which is similar to the surface tension of the liquid film.
According to this analysis, the shear stress at any point on the cross-section of the torsion bar, along any direction, is equal to the slope of the film in the vertical direction at that point.
It can be seen that the maximum shear stress in the cross section of the torsion bar is equal to the maximum slope of the membrane. However, it should be noted that the direction of maximum shear stress is perpendicular to the direction of maximum slope.
By making this assumption, it is possible to determine the maximum shear stress and relative torsion angle of the straight bar of non-circular section listed in Table 1 below.
Fig. 5 Membrane analogy model
3. Calculation of torsional shear stress and twist 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 deformation;
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 twist end, R is the outer radius of the circle and L is the spacing between two sections).
Fig. 6 Schematic diagram of bar torsion with solid circular section
The shear stress in ρ in the circular section is:
Under the same torque condition, the shear stress (τ) in a bar of circular cross-section is proportional to the distance from the center of the section (ρ). This means that the greater the distance from the center, the greater the shear stress.
When the distance from the center is equal to the radius (R) of the circular section, the maximum shear stress is obtained at the edge.
The modulus of the torsional section (Wp) of a circular shaft can be expressed as IP/R, where IP is the polar moment of inertia. This value is related only to the geometric dimensions of the section and not to the cross-sectional area.
The maximum shear stress (τ max) can be calculated as T/WP, where T is the applied torque.
For a solid shaft with a circular section, the torsional section modulus (WP) is approximately equal to 0.2 times the cube of the diameter (D).
The twist angle (φ) of a round bar under torsion is related to the torsional stiffness (GIP) of the circular section, which reflects the ability of the shaft to resist deformation.
The relative twist angles of two sections a distance L can be calculated using a twist formula.
Relative angle of twist:
Circular shaft stiffness condition:
(2) Hollow circular shaft
The torsion coefficient of the hollow circular shaft section 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 resistance decreases by 40%, but under the same material and length the weight difference is 2.8 times.
(3) Thin-wall closed tube
A round tube with a wall thickness (a) much smaller than its radius (R0) – normally considered ≤ R0/10 – is known as a thin-walled round tube. This type of tube can have any shape and the same section.
As it is a thin-walled tube, it is assumed that the shear stress is uniformly distributed throughout the entire wall thickness.
